\displaystyle \text{Using integration, find the area of the following regions:}

\displaystyle \textbf{Question 1: }~\text{Calculate the area of the region bounded by the parabolas } \\ y^{2}=6x\text{ and }x^{2}=6y.
\displaystyle \text{Answer:} \displaystyle y^2=x\text{ is a parabola opening sideways with vertex at }O(0,0)\text{ and the positive } \\ x\text{-axis as axis of symmetry}
\displaystyle x^2=y\text{ is a parabola opening upwards with vertex at }O(0,0)\text{ and the positive } \\ y\text{-axis as axis of symmetry}
\displaystyle \text{Solving the two equations,}
\displaystyle x^2=y
\displaystyle y^2=x\Rightarrow x^2=y^4
\displaystyle \Rightarrow y^4-y=0
\displaystyle \Rightarrow y(y^3-1)=0
\displaystyle \Rightarrow y=0\text{ or }y=1
\displaystyle \text{Correspondingly, }x=0\text{ or }x=1
\displaystyle \text{Thus, }O(0,0)\text{ and }A(1,1)\text{ are the points of intersection of the curves}
\displaystyle \text{Consider a vertical strip of length }|y_2-y_1|\text{ and width }dx
\displaystyle \text{Area of the approximating rectangle }=|y_2-y_1|\,dx
\displaystyle \text{The strip moves from }x=0\text{ to }x=1
\displaystyle \text{Area of the shaded region }=\int_{0}^{1}|y_2-y_1|\,dx
\displaystyle \text{Since }y_2-y_1>0,\ |y_2-y_1|=y_2-y_1
\displaystyle \text{Here }y_2=\sqrt{x}\text{ and }y_1=x^2
\displaystyle A=\int_{0}^{1}(\sqrt{x}-x^2)\,dx
\displaystyle =\left[\frac{2}{3}x^{3/2}-\frac{x^3}{3}\right]_{0}^{1}
\displaystyle =\left(\frac{2}{3}-\frac{1}{3}\right)
\displaystyle =\frac{1}{3}
\displaystyle \text{Hence, the area enclosed by the curves is }\frac{1}{3}\text{ sq. units}

\displaystyle \textbf{Question 2: }~\text{Find the area of the region common to the parabolas } \\ 4y^{2}=9x\text{ and }3x^{2}=16y.
\displaystyle \text{Answer:} \displaystyle 3x^2=16y\text{ represents a parabola with vertex at }(0,0)\text{ opening upwards and } \\ \text{symmetric about the positive }y\text{-axis}
\displaystyle 4y^2=9x\text{ represents a parabola with vertex at }(0,0)\text{ opening sideways and } \\ \text{symmetric about the positive }x\text{-axis}
\displaystyle \text{Solving the equations }3x^2=16y\text{ and }4y^2=9x
\displaystyle y=\frac{3x^2}{16}
\displaystyle 4\left(\frac{3x^2}{16}\right)^2=9x
\displaystyle \Rightarrow \frac{9x^4}{64}=9x
\displaystyle \Rightarrow x(x^3-64)=0
\displaystyle \Rightarrow x=0\text{ or }x=4
\displaystyle \text{Correspondingly, }y=0\text{ or }y=3
\displaystyle \text{Thus, the points of intersection are }O(0,0)\text{ and }A(4,3)
\displaystyle \text{Consider a vertical strip of length }|y_2-y_1|\text{ and width }dx
\displaystyle \text{Area of the approximating rectangle }=|y_2-y_1|\,dx
\displaystyle \text{The strip moves from }x=0\text{ to }x=4
\displaystyle \text{Area of the shaded region }=\int_{0}^{4}|y_2-y_1|\,dx
\displaystyle \text{Since }y_2-y_1>0,\ |y_2-y_1|=y_2-y_1
\displaystyle \text{From }4y^2=9x,\ y_2=\frac{3}{2}\sqrt{x}
\displaystyle \text{From }3x^2=16y,\ y_1=\frac{3x^2}{16}
\displaystyle A=\int_{0}^{4}\left(\frac{3}{2}\sqrt{x}-\frac{3x^2}{16}\right)dx
\displaystyle =\frac{3}{2}\int_{0}^{4}x^{1/2}dx-\frac{3}{16}\int_{0}^{4}x^2dx
\displaystyle =\frac{3}{2}\left[\frac{x^{3/2}}{3/2}\right]_{0}^{4}-\frac{3}{16}\left[\frac{x^3}{3}\right]_{0}^{4}
\displaystyle =4^{3/2}-\frac{1}{16}\times4^3
\displaystyle =8-4
\displaystyle =4
\displaystyle \text{Hence, the area bounded by the two curves is }4\text{ sq. units}

\displaystyle \textbf{Question 3: }~\text{Find the area of the region bounded by }y=\sqrt{x}\text{ and }y=x.
\displaystyle \text{Answer:} \displaystyle y=\sqrt{x}\text{ represents a parabola opening sideways with vertex at }O(0,0)\text{ and the positive } \\ x\text{-axis as axis of symmetry}
\displaystyle x=y\text{ is a straight line passing through }O(0,0)\text{ and making an angle }45^\circ\text{ with the }x\text{-axis}
\displaystyle \text{Solving the equations }y=\sqrt{x}\text{ and }x=y
\displaystyle y^2=x=y
\displaystyle \Rightarrow y^2=y
\displaystyle \Rightarrow y(y-1)=0
\displaystyle \Rightarrow y=0\text{ or }y=1
\displaystyle \text{Correspondingly, }x=0\text{ or }x=1
\displaystyle \text{Thus, the line intersects the parabola at }O(0,0)\text{ and }A(1,1)
\displaystyle \text{Consider a vertical strip of length }|y_2-y_1|\text{ and width }dx
\displaystyle \text{Area of the approximating rectangle }=|y_2-y_1|\,dx
\displaystyle \text{The strip moves from }x=0\text{ to }x=1
\displaystyle \text{Area of the shaded region }=\int_{0}^{1}|y_2-y_1|\,dx
\displaystyle \text{Since }y_2>y_1,\ |y_2-y_1|=y_2-y_1
\displaystyle \text{Here }y_2=\sqrt{x}\text{ and }y_1=x
\displaystyle A=\int_{0}^{1}(\sqrt{x}-x)\,dx
\displaystyle =\left[\frac{2}{3}x^{3/2}-\frac{x^2}{2}\right]_{0}^{1}
\displaystyle =\frac{2}{3}-\frac{1}{2}
\displaystyle =\frac{1}{6}
\displaystyle \text{Hence, the area bounded by the parabola and the straight line is }\frac{1}{6}\text{ sq. units}

\displaystyle \textbf{Question 4: }~\text{Find the area bounded by the curve } \\ y=4-x^{2}\text{ and the lines }y=0,\ y=3.
\displaystyle \text{Answer:} \displaystyle y = 4 - x^2 \text{ is a parabola with vertex } (0,4), \text{ opening downwards and having } \\ \text{axis of symmetry the } y\text{-axis}
\displaystyle y = 0 \text{ is the } x\text{-axis cutting the parabola at } A(2,0) \text{ and } A'(-2,0)
\displaystyle y = 3 \text{ is a line parallel to the } x\text{-axis cutting the parabola at } B(1,3) \text{ and } B'(-1,3)
\displaystyle \text{It cuts the } y\text{-axis at } C(0,3)
\displaystyle \text{Required area is the shaded area } ABB'A' = 2(\text{area } ABCO)
\displaystyle \text{Consider a horizontal strip of length } |x_2 - x_1| \text{ and width } dy
\displaystyle \text{Area of approximating rectangle } = |x_2 - x_1|\,dy
\displaystyle \text{The strip moves from } y = 0 \text{ to } y = 3
\displaystyle \text{Area of shaded region } = 2\int_0^3 |x_2 - x_1|\,dy
\displaystyle \text{Since } x_2 > x_1,\ |x_2 - x_1| = x_2 - x_1
\displaystyle \text{From } y = 4 - x^2,\ x = \sqrt{4 - y}
\displaystyle x_2 = \sqrt{4 - y},\ x_1 = 0
\displaystyle A = 2\int_0^3 (\sqrt{4 - y} - 0)\,dy
\displaystyle A = 2\int_0^3 \sqrt{4 - y}\,dy
\displaystyle A = 2\left[-\frac{2}{3}(4 - y)^{3/2}\right]_0^3
\displaystyle A = 2\left[\frac{2}{3}(4^{3/2} - 1^{3/2})\right]
\displaystyle A = 2 \times \frac{2}{3}(8 - 1)
\displaystyle A = \frac{4}{3} \times 7
\displaystyle A = \frac{28}{3} \text{ sq. units}

\displaystyle \textbf{Question 5: }~\text{Find the area of the region }\{(x,y):\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\le 1\le \frac{x}{a}+\frac{y}{b}\}.
\displaystyle \text{Answer:} \displaystyle \text{Let } R = \{(x,y): \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1 \le \frac{x}{a} + \frac{y}{b}\}
\displaystyle \Rightarrow R_1 = \{(x,y): \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1\}
\displaystyle \text{and } R_2 = \{(x,y): 1 \le \frac{x}{a} + \frac{y}{b}\}
\displaystyle \text{Then } R = R_1 \cap R_2
\displaystyle \text{Consider } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\displaystyle \text{This represents an ellipse symmetrical about both axes}
\displaystyle \text{It cuts the } x\text{-axis at } A(a,0), A'(-a,0)
\displaystyle \text{and the } y\text{-axis at } B(0,b), B'(0,-b)
\displaystyle \Rightarrow R_1 = \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1\right) \text{ represents the region inside the ellipse}
\displaystyle \text{Now consider } \frac{x}{a} + \frac{y}{b} = 1
\displaystyle \text{This represents a straight line cutting the } x\text{-axis at } (a,0)
\displaystyle \text{and the } y\text{-axis at } (0,b)
\displaystyle \Rightarrow R_2 = \left(\frac{x}{a} + \frac{y}{b} \ge 1\right) \text{ represents the region above the line}
\displaystyle \text{Hence } R = R_1 \cap R_2 \text{ is the region bounded by the ellipse and the line}
\displaystyle \text{In the shaded region, consider a vertical strip of width } dx
\displaystyle \text{Let } P(x,y_2) \text{ lie on the ellipse and } Q(x,y_1) \text{ on the line}
\displaystyle \text{Length of strip } = |y_2 - y_1|
\displaystyle \text{Area of strip } = |y_2 - y_1|\,dx
\displaystyle \text{The strip moves from } x=0 \text{ to } x=a
\displaystyle \text{Since } y_2 > y_1,\ |y_2 - y_1| = y_2 - y_1
\displaystyle \text{Equation of ellipse gives } y_2 = \frac{b}{a}\sqrt{a^2 - x^2}
\displaystyle \text{Equation of line gives } y_1 = \frac{b}{a}(a - x)
\displaystyle \text{Area } A = \int_0^a (y_2 - y_1)\,dx
\displaystyle A = \int_0^a \left(\frac{b}{a}\sqrt{a^2 - x^2} - \frac{b}{a}(a - x)\right) dx
\displaystyle A = \frac{b}{a}\int_0^a \sqrt{a^2 - x^2}\,dx - \frac{b}{a}\int_0^a (a - x)\,dx
\displaystyle A = \frac{b}{a}\left[\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)\right]_0^a - \frac{b}{a}\left[ax - \frac{x^2}{2}\right]_0^a
\displaystyle A = \frac{b}{a}\left[\frac{a^2}{2}\cdot\frac{\pi}{2}\right] - \frac{b}{a}\left[a^2 - \frac{a^2}{2}\right]
\displaystyle A = \frac{b}{a}\left(\frac{\pi a^2}{4} - \frac{a^2}{2}\right)
\displaystyle A = \frac{ab}{4}(\pi - 2) \text{ sq. units}

\displaystyle \textbf{Question 6: }~\text{Using integration, find the area of the region bounded by the} \\ \text{triangle whose vertices are }(2,1),\ (3,4)\text{ and }(5,2).
\displaystyle \text{Answer:} \displaystyle \text{Consider the points } A(2,1), B(3,4) \text{ and } C(5,2)
\displaystyle \text{We need to find the area of shaded triangle } ABC
\displaystyle \text{Equation of } AB \text{ is}
\displaystyle y - 1 = \frac{4 - 1}{3 - 2}(x - 2)
\displaystyle \Rightarrow y = 3x - 5
\displaystyle \Rightarrow 3x - y - 5 = 0 \quad (1)
\displaystyle \text{Equation of } BC \text{ is}
\displaystyle y - 4 = \frac{2 - 4}{5 - 3}(x - 3)
\displaystyle \Rightarrow y = -x + 7
\displaystyle \Rightarrow x + y - 7 = 0 \quad (2)
\displaystyle \text{Equation of } CA \text{ is}
\displaystyle y - 2 = \frac{1 - 2}{2 - 5}(x - 5)
\displaystyle \Rightarrow y = \frac{x + 1}{3}
\displaystyle \Rightarrow x - 3y + 1 = 0 \quad (3)
\displaystyle \text{Area of } \triangle ABC = \text{Area of } \triangle ABD + \text{Area of } \triangle DBC
\displaystyle \text{In } \triangle ABD, \text{ consider a vertical strip}
\displaystyle \text{Let } P(x,y_2) \text{ lie on } AB \text{ and } Q(x,y_1) \text{ lie on } AD
\displaystyle \text{Area of approximating rectangle } = |y_2 - y_1|\,dx
\displaystyle \text{The strip moves from } x = 2 \text{ to } x = 3
\displaystyle \text{Since } y_2 > y_1,\ |y_2 - y_1| = y_2 - y_1
\displaystyle y_2 = 3x - 5,\quad y_1 = \frac{x + 1}{3}
\displaystyle \text{Area of } \triangle ABD = \int_2^3 \left[(3x - 5) - \frac{x + 1}{3}\right] dx
\displaystyle A = \int_2^3 \frac{9x - 15 - x - 1}{3}\,dx
\displaystyle A = \int_2^3 \frac{8x - 16}{3}\,dx
\displaystyle A = \frac{1}{3}\left[4x^2 - 16x\right]_2^3
\displaystyle A = \frac{1}{3}[(36 - 48) - (16 - 32)]
\displaystyle A = \frac{1}{3}(4)
\displaystyle A = \frac{4}{3} \text{ sq. units}
\displaystyle \text{Similarly, in } \triangle DBC, \text{ consider a vertical strip}
\displaystyle \text{Let } S(x,y_4) \text{ lie on } BC \text{ and } R(x,y_3) \text{ lie on } AC
\displaystyle \text{Area of approximating rectangle } = |y_4 - y_3|\,dx
\displaystyle \text{The strip moves from } x = 3 \text{ to } x = 5
\displaystyle \text{Since } y_4 > y_3,\ |y_4 - y_3| = y_4 - y_3
\displaystyle y_4 = 7 - x,\quad y_3 = \frac{x + 1}{3}
\displaystyle \text{Area of } \triangle DBC = \int_3^5 \left[(7 - x) - \frac{x + 1}{3}\right] dx
\displaystyle A = \int_3^5 \frac{21 - 3x - x - 1}{3}\,dx
\displaystyle A = \int_3^5 \frac{20 - 4x}{3}\,dx
\displaystyle A = \frac{1}{3}\left[20x - 2x^2\right]_3^5
\displaystyle A = \frac{1}{3}[(100 - 50) - (60 - 18)]
\displaystyle A = \frac{1}{3}(8)
\displaystyle A = \frac{8}{3} \text{ sq. units}
\displaystyle \text{Area of } \triangle ABC = \frac{4}{3} + \frac{8}{3}
\displaystyle \text{Area of } \triangle ABC = 4 \text{ sq. units}

\displaystyle \textbf{Question 7: }~\text{Using integration, find the area of the region bounded by the } \\ \text{triangle }ABC\text{ whose vertices }A,B,C\text{ are }(-1,1),\ (0,5)\text{ and }(3,2)\text{ respectively.}
\displaystyle \text{Answer:} \displaystyle \text{Equation of line } AB \text{ is}
\displaystyle y - 1 = \frac{5 - 1}{0 - (-1)}(x - (-1))
\displaystyle \Rightarrow y = 4x + 5
\displaystyle \text{Area under the line } AB = \text{Area } ABDO
\displaystyle = \int_{-1}^{0} (4x + 5)\,dx
\displaystyle = \left[2x^2 + 5x\right]_{-1}^{0}
\displaystyle = 0 - (2 - 5)
\displaystyle \Rightarrow \text{Area } ABDO = 3 \text{ sq. units } \quad (1)
\displaystyle \text{Equation of line } BC \text{ is}
\displaystyle y - 5 = \frac{2 - 5}{3 - 0}(x - 0)
\displaystyle \Rightarrow y = -x + 5
\displaystyle \text{Area under line } BC = \text{Area } OBCP
\displaystyle = \int_{0}^{3} (-x + 5)\,dx
\displaystyle = \left[-\frac{x^2}{2} + 5x\right]_{0}^{3}
\displaystyle = -\frac{9}{2} + 15
\displaystyle \Rightarrow \text{Area } OBCP = \frac{21}{2} \text{ sq. units } \quad (2)
\displaystyle \text{Equation of line } CA \text{ is}
\displaystyle y - 2 = \frac{1 - 2}{-1 - 3}(x - 3)
\displaystyle \Rightarrow 4y = x + 5
\displaystyle \text{Area under line } AC = \text{Area } ACPA
\displaystyle A = \int_{-1}^{3} \frac{x + 5}{4}\,dx
\displaystyle A = \frac{1}{4}\left[\frac{x^2}{2} + 5x\right]_{-1}^{3}
\displaystyle A = \frac{1}{4}\left[\frac{9}{2} + 15 - \frac{1}{2} + 5\right]
\displaystyle \Rightarrow \text{Area } ACPA = \frac{24}{4} = 6 \text{ sq. units } \quad (3)
\displaystyle \text{From } (1), (2) \text{ and } (3)
\displaystyle \text{Area } \triangle ABC = \text{Area } ABDO + \text{Area } OBCP - \text{Area } ACPA
\displaystyle A = 3 + \frac{21}{2} - 6
\displaystyle A = \frac{21}{2} - 3
\displaystyle A = \frac{15}{2} \text{ sq. units}

\displaystyle \textbf{Question 8: }~\text{Using integration, find the area of the triangular region, the equations } \\ \text{of whose sides are }y=2x+1,\ y=3x+1\text{ and }x=4.
\displaystyle \text{Answer:} \displaystyle \text{Solving the given equations}
\displaystyle \text{The points of intersection of the three lines are } A(0,1), B(4,13) \text{ and } C(4,9)
\displaystyle \text{We need to find the area of } \triangle ABC
\displaystyle \text{Area under line } AB = \text{Area } OABCL
\displaystyle \text{Equation of } AB \text{ is } y = 3x + 1
\displaystyle \text{As } x \text{ moves from } A(0,1) \text{ to } B(4,13),\ x \text{ varies from } 0 \text{ to } 4
\displaystyle \text{Area } OABCL = \int_0^4 (3x + 1)\,dx
\displaystyle = \left[\frac{3x^2}{2} + x\right]_0^4
\displaystyle = \left[\frac{3(4)^2}{2} + 4\right]
\displaystyle = 24 + 4 = 28 \text{ sq. units}
\displaystyle \text{Area under line } AC = \text{Area } OACL
\displaystyle \text{Equation of } AC \text{ is } y = 2x + 1
\displaystyle \text{As } x \text{ moves from } A(0,1) \text{ to } C(4,9),\ x \text{ varies from } 0 \text{ to } 4
\displaystyle \text{Area } OACL = \int_0^4 (2x + 1)\,dx
\displaystyle = \left[x^2 + x\right]_0^4
\displaystyle = 16 + 4 = 20 \text{ sq. units}
\displaystyle \text{Area of } \triangle ABC = \text{Area } OABCL - \text{Area } OACL
\displaystyle \text{Area of } \triangle ABC = 28 - 20
\displaystyle \text{Area of } \triangle ABC = 8 \text{ sq. units}
\displaystyle \text{Area of triangle formed by the three given lines } = 8 \text{ sq. units}

\displaystyle \textbf{Question 9: }~\text{Find the area of the region }\{(x,y):y^{2}\le 8x,\ x^{2}+y^{2}\le 9\}.
\displaystyle \text{Answer:} \displaystyle \text{Let } R = \{(x,y): y^2 \le 8x,\ x^2 + y^2 \le 9\}
\displaystyle R_1 = \{(x,y): y^2 \le 8x\}
\displaystyle R_2 = \{(x,y): x^2 + y^2 \le 9\}
\displaystyle \text{Thus } R = R_1 \cap R_2
\displaystyle \text{Now } y^2 = 8x \text{ represents a parabola with vertex } O(0,0) \text{ symmetric about the } x\text{-axis}
\displaystyle \text{Thus } R_1 \text{ represents the region inside the parabola}
\displaystyle \text{Also } x^2 + y^2 = 9 \text{ represents a circle with centre } O(0,0) \text{ and radius } 3
\displaystyle \text{The circle cuts the } x\text{-axis at } (3,0), (-3,0) \text{ and the } y\text{-axis at } (0,3), (0,-3)
\displaystyle \text{Thus } R_2 \text{ represents the region inside the circle}
\displaystyle \Rightarrow R = R_1 \cap R_2 = \text{Area } OACA'O = 2(\text{shaded area } OACO) \quad (1)
\displaystyle \text{The points of intersection are obtained by solving } y^2 = 8x \text{ and } x^2 + y^2 = 9
\displaystyle \Rightarrow x^2 + 8x = 9
\displaystyle \Rightarrow x^2 + 8x - 9 = 0
\displaystyle \Rightarrow (x + 9)(x - 1) = 0
\displaystyle \Rightarrow x = -9 \text{ or } x = 1
\displaystyle \text{Since the parabola lies for } x \ge 0,\ x = 1 \text{ is admissible}
\displaystyle \Rightarrow y^2 = 8
\displaystyle \Rightarrow y = \pm 2\sqrt{2}
\displaystyle \text{Thus the points of intersection are } A(1,2\sqrt{2}) \text{ and } A'(1,-2\sqrt{2})
\displaystyle \text{Area } OACO = \text{Area } OADO + \text{Area } DACD \quad (2)
\displaystyle \text{Area } OADO = \int_0^1 \sqrt{8x}\,dx \quad [\text{area under } y^2 = 8x]
\displaystyle = 2\sqrt{2}\int_0^1 x^{1/2}\,dx
\displaystyle = 2\sqrt{2}\left[\frac{x^{3/2}}{3/2}\right]_0^1
\displaystyle \Rightarrow \text{Area } OADO = \frac{4\sqrt{2}}{3} \quad (3)
\displaystyle \text{Area } DACD = \int_1^3 \sqrt{9 - x^2}\,dx
\displaystyle = \left[\frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right)\right]_1^3
\displaystyle = 0 + \frac{9}{2}\sin^{-1}(1) - \frac{1}{2}\sqrt{8} - \frac{9}{2}\sin^{-1}\left(\frac{1}{3}\right)
\displaystyle = \frac{9\pi}{4} - \sqrt{2} - \frac{9}{2}\sin^{-1}\left(\frac{1}{3}\right) \quad (4)
\displaystyle \text{From } (1), (2), (3) \text{ and } (4)
\displaystyle \text{Area } R = 2\left(\frac{4\sqrt{2}}{3} + \frac{9\pi}{4} - \sqrt{2} - \frac{9}{2}\sin^{-1}\left(\frac{1}{3}\right)\right)
\displaystyle \text{Area } R = 2\left(\frac{\sqrt{2}}{3} + \frac{9\pi}{4} - \frac{9}{2}\sin^{-1}\left(\frac{1}{3}\right)\right) \text{ sq. units}

\displaystyle \textbf{Question 10: }~\text{Find the area of the region common to the circle } \\ x^{2}+y^{2}=16\text{ and the parabola }y^{2}=6x.
\displaystyle \text{Answer:} \displaystyle \text{Points of intersection of the parabola and the circle are obtained by solving}
\displaystyle x^2 + y^2 = 16 \text{ and } y^2 = 6x
\displaystyle \Rightarrow x^2 + 6x = 16
\displaystyle \Rightarrow x^2 + 6x - 16 = 0
\displaystyle \Rightarrow (x + 8)(x - 2) = 0
\displaystyle \Rightarrow x = -8 \text{ or } x = 2
\displaystyle \text{Since the parabola lies for } x \ge 0,\ x = 2 \text{ is admissible}
\displaystyle \Rightarrow y^2 = 6 \times 2 = 12
\displaystyle \Rightarrow y = \pm 2\sqrt{3}
\displaystyle \text{Thus } B(2,2\sqrt{3}) \text{ and } B'(2,-2\sqrt{3}) \text{ are the points of intersection}
\displaystyle \text{Now, required area } = \text{area } (OBAB'O)
\displaystyle = 2(\text{area } OBAO)
\displaystyle = 2\{\text{area } OBDO + \text{area } DBAD\}
\displaystyle = 2\left[\int_0^2 \sqrt{6x}\,dx + \int_2^4 \sqrt{16 - x^2}\,dx\right]
\displaystyle = 2\left[\int_0^2 \sqrt{6}\,x^{1/2}\,dx + \int_2^4 \sqrt{16 - x^2}\,dx\right]
\displaystyle = 2\left[\sqrt{6}\left(\frac{2}{3}x^{3/2}\right)_0^2 + \left(\frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right)\right)_2^4\right]
\displaystyle = 2\left[\sqrt{6}\cdot\frac{2}{3}\cdot(2\sqrt{2}) + \left(0 + 8\sin^{-1}(1) - \sqrt{12} - 8\sin^{-1}\left(\frac{1}{2}\right)\right)\right]
\displaystyle = 2\left[\frac{8\sqrt{3}}{3} + 8\left(\frac{\pi}{2} - \frac{\pi}{6}\right) - 2\sqrt{3}\right]
\displaystyle = 2\left[\frac{8\sqrt{3} - 6\sqrt{3}}{3} + \frac{8\pi}{3}\right]
\displaystyle = 2\left[\frac{2\sqrt{3}}{3} + \frac{8\pi}{3}\right]
\displaystyle = \frac{4\sqrt{3}}{3} + \frac{16\pi}{3} \text{ sq. units}

\displaystyle \textbf{Question 11: }~\text{Find the area of the region between the circles }x^{2}+y^{2}=4 \\ \text{ and }(x-2)^{2}+y^{2}=4.
\displaystyle \text{Answer:} \displaystyle x^2 + y^2 = 4 \quad (1) \text{ represents a circle with centre } O(0,0) \text{ and radius } 2
\displaystyle (x - 2)^2 + y^2 = 4 \quad (2) \text{ represents a circle with centre } A(2,0) \text{ and radius } 2
\displaystyle \text{Points of intersection of the two circles are obtained by solving } (1) \text{ and } (2)
\displaystyle (x - 2)^2 + y^2 = x^2 + y^2
\displaystyle \Rightarrow x^2 - 4x + 4 = x^2
\displaystyle \Rightarrow x = 1
\displaystyle \Rightarrow y^2 = 4 - 1 = 3
\displaystyle \Rightarrow y = \pm \sqrt{3}
\displaystyle \text{Thus } B(1,\sqrt{3}) \text{ and } B'(1,-\sqrt{3}) \text{ are the points of intersection}
\displaystyle \text{We need to find shaded area } = 2 \times \text{area } (OBAO) \quad (3)
\displaystyle \text{Area } (OBAO) = \text{area } (OBPO) + \text{area } (BPAP)
\displaystyle \text{Consider a vertical strip of width } dx
\displaystyle \text{Since } y > 0,\ |y_1| = y_1 \text{ and } |y_2| = y_2
\displaystyle \text{Area } (OBAO) = \int_0^1 y_1\,dx + \int_1^2 y_2\,dx
\displaystyle = \int_0^1 \sqrt{4 - (x - 2)^2}\,dx + \int_1^2 \sqrt{4 - x^2}\,dx
\displaystyle = \left[\frac{1}{2}(x - 2)\sqrt{4 - (x - 2)^2} + 2\sin^{-1}\left(\frac{x - 2}{2}\right)\right]_0^1
\displaystyle \quad + \left[\frac{1}{2}x\sqrt{4 - x^2} + 2\sin^{-1}\left(\frac{x}{2}\right)\right]_1^2
\displaystyle = \left[-\frac{\sqrt{3}}{2} + 2\sin^{-1}\left(-\frac{1}{2}\right)\right] - \left[0 + 2\sin^{-1}(-1)\right]
\displaystyle \quad + \left[0 + 2\sin^{-1}(1)\right] - \left[\frac{\sqrt{3}}{2} + 2\sin^{-1}\left(\frac{1}{2}\right)\right]
\displaystyle = -\frac{\sqrt{3}}{2} - 2\sin^{-1}\left(\frac{1}{2}\right) + \pi - \frac{\sqrt{3}}{2} - 2\sin^{-1}\left(\frac{1}{2}\right)
\displaystyle = -\sqrt{3} - 4\sin^{-1}\left(\frac{1}{2}\right) + \pi
\displaystyle = -\sqrt{3} - \frac{4\pi}{6} + \pi
\displaystyle = \frac{4\pi}{3} - \sqrt{3}
\displaystyle \text{From } (3),\ \text{Shaded area } = 2\left(\frac{4\pi}{3} - \sqrt{3}\right)
\displaystyle \text{Shaded area } = \frac{8\pi}{3} - 2\sqrt{3} \text{ sq. units}

\displaystyle \textbf{Question 12: }~\text{Find the area of the region included between the parabola } \\ y^{2}=x\text{ and the line }x+y=2.
\displaystyle \text{Answer:} \displaystyle \text{We have } y^2 = x \text{ and } x + y = 2
\displaystyle \text{To find the points of intersection, we solve both equations}
\displaystyle y^2 + y - 2 = 0
\displaystyle \Rightarrow (y + 2)(y - 1) = 0
\displaystyle \Rightarrow y = -2 \text{ or } y = 1
\displaystyle \Rightarrow x = y^2 = 4 \text{ or } 1
\displaystyle \text{Consider a horizontal strip of length } |x_2 - x_1| \text{ and width } dy
\displaystyle \text{Let } P(x_2,y) \text{ lie on the line and } Q(x_1,y) \text{ lie on the parabola}
\displaystyle \text{Area of approximating rectangle } = |x_2 - x_1|\,dy
\displaystyle \text{The strip moves from } y = -2 \text{ to } y = 1
\displaystyle \text{Required area } = \text{area } (OADO) = \int_{-2}^{1} |x_2 - x_1|\,dy
\displaystyle \text{Since } x_2 > x_1,\ |x_2 - x_1| = x_2 - x_1
\displaystyle x_2 = 2 - y,\quad x_1 = y^2
\displaystyle \text{Area } = \int_{-2}^{1} \left[(2 - y) - y^2\right] dy
\displaystyle = \int_{-2}^{1} (2 - y - y^2)\,dy
\displaystyle = \left[2y - \frac{y^2}{2} - \frac{y^3}{3}\right]_{-2}^{1}
\displaystyle = \left(2 - \frac{1}{2} - \frac{1}{3}\right) - \left(-4 - 2 + \frac{8}{3}\right)
\displaystyle = \frac{7}{6} + \frac{10}{6}
\displaystyle = \frac{9}{2} \text{ sq. units}
\displaystyle \text{Area enclosed by the line and the parabola } = \frac{9}{2} \text{ sq. units}

\displaystyle \textbf{Question 13: }~\text{Draw a rough sketch of the region }\{(x,y):y^{2}\le 3x,\ 3x^{2}+3y^{2}\le 16\}\text{ and find the area enclosed by the region using method of integration.}
\displaystyle \text{Answer:} \displaystyle \text{The given region is the intersection of } y^2 \le 3x \text{ and } 3x^2 + 3y^2 \le 16
\displaystyle \text{Now } y^2 = 3x \text{ is a parabola with vertex at } (0,0) \text{ opening along the positive } x\text{-axis}
\displaystyle \text{Also } 3x^2 + 3y^2 = 16 \text{ represents a circle with centre at origin}
\displaystyle \text{The radius of the circle is } \sqrt{\frac{16}{3}}
\displaystyle \text{Corresponding equations of the curves are}
\displaystyle y^2 = 3x \quad (1)
\displaystyle 3x^2 + 3y^2 = 16 \quad (2)
\displaystyle \text{Substituting } y^2 = 3x \text{ from } (1) \text{ into } (2)
\displaystyle 3x^2 + 9x = 16
\displaystyle \Rightarrow 3x^2 + 9x - 16 = 0
\displaystyle \Rightarrow x = \frac{-9 \pm \sqrt{81 + 192}}{6}
\displaystyle \Rightarrow x = \frac{-9 \pm \sqrt{273}}{6}
\displaystyle \text{Since the region lies in } x \ge 0,\ x = \frac{-9 + \sqrt{273}}{6} \text{ is admissible}
\displaystyle \text{Let } a = \frac{-9 + \sqrt{273}}{6}
\displaystyle \text{The required area } = 2(\text{Area } OACO + \text{Area } CABC)
\displaystyle \text{Consider a vertical strip of width } dx
\displaystyle \text{Area of } OACO = \int_0^{a} |y_1|\,dx
\displaystyle \text{Since } y_1 > 0,\ |y_1| = y_1
\displaystyle y_1 = \sqrt{3x}
\displaystyle \text{Area } OACO = \int_0^{a} \sqrt{3x}\,dx
\displaystyle = \sqrt{3}\int_0^{a} x^{1/2}\,dx
\displaystyle = \sqrt{3}\left[\frac{2}{3}x^{3/2}\right]_0^{a}
\displaystyle = \frac{2\sqrt{3}}{3}a^{3/2}
\displaystyle \text{Similarly, area } CABC \text{ is bounded by the circle}
\displaystyle \text{Area } CABC = \int_a^{\sqrt{16/3}} |y_2|\,dx
\displaystyle \text{Since } y_2 > 0,\ |y_2| = y_2
\displaystyle y_2 = \sqrt{\frac{16}{3} - x^2}
\displaystyle \text{Area } CABC = \int_a^{\sqrt{16/3}} \sqrt{\frac{16}{3} - x^2}\,dx
\displaystyle = \left[\frac{x}{2}\sqrt{\frac{16}{3} - x^2} + \frac{8}{3}\sin^{-1}\left(\frac{x\sqrt{3}}{4}\right)\right]_a^{\sqrt{16/3}}
\displaystyle = 0 + \frac{8}{3}\sin^{-1}(1) - \frac{a}{2}\sqrt{\frac{16}{3} - a^2} - \frac{8}{3}\sin^{-1}\left(\frac{a\sqrt{3}}{4}\right)
\displaystyle = \frac{4\pi}{3} - \frac{a}{2}\sqrt{\frac{16}{3} - a^2} - \frac{8}{3}\sin^{-1}\left(\frac{a\sqrt{3}}{4}\right)
\displaystyle \text{Hence required area } = 2\left[\frac{2\sqrt{3}}{3}a^{3/2} + \frac{4\pi}{3} - \frac{a}{2}\sqrt{\frac{16}{3} - a^2} - \frac{8}{3}\sin^{-1}\left(\frac{a\sqrt{3}}{4}\right)\right]
\displaystyle \text{Required area } = \frac{4\sqrt{3}}{3}a^{3/2} + \frac{8\pi}{3} - a\sqrt{\frac{16}{3} - a^2} - \frac{16}{3}\sin^{-1}\left(\frac{a\sqrt{3}}{4}\right)
\displaystyle \text{where } a = \frac{-9 + \sqrt{273}}{6}

\displaystyle \textbf{Question 14: }~\text{Draw a rough sketch of the region }\{(x,y):y^{2}\le 5x,\ 5x^{2}+5y^{2}\le 36\}\text{ and find the area enclosed by the region using method of integration.}
\displaystyle \text{Answer:}

\displaystyle \text{The given region is the intersection of } y^2 \le 5x \text{ and } 5x^2 + 5y^2 \le 36
\displaystyle \text{Clearly } y^2 = 5x \text{ is a parabola with vertex at the origin and axis along the } x\text{-axis opening in the positive direction}
\displaystyle \text{Also } 5x^2 + 5y^2 \le 36 \text{ represents a circle with centre at the origin and radius } \sqrt{\frac{36}{5}} = \frac{6}{\sqrt{5}}
\displaystyle \text{Corresponding equations of the curves are}
\displaystyle y^2 = 5x \quad (1)
\displaystyle 5x^2 + 5y^2 = 36 \quad (2)
\displaystyle \text{Substituting } y^2 = 5x \text{ from } (1) \text{ into } (2)
\displaystyle 5x^2 + 25x = 36
\displaystyle \Rightarrow 5x^2 + 25x - 36 = 0
\displaystyle \Rightarrow x = \frac{-25 \pm \sqrt{625 + 720}}{10}
\displaystyle \Rightarrow x = \frac{-25 \pm \sqrt{1345}}{10}
\displaystyle \text{From the figure, } x \ge 0 \text{ for the required region}
\displaystyle \Rightarrow x = \frac{-25 + \sqrt{1345}}{10}
\displaystyle \text{Let } a = \frac{-25 + \sqrt{1345}}{10}
\displaystyle \text{The required area } A = 2(\text{Area } OACO + \text{Area } CABC)
\displaystyle \text{Consider a vertical strip of width } dx
\displaystyle \text{Area of } OACO = \int_0^{a} |y_1|\,dx
\displaystyle \text{Since } y_1 > 0,\ |y_1| = y_1
\displaystyle y_1 = \sqrt{5x}
\displaystyle \text{Area } OACO = \int_0^{a} \sqrt{5x}\,dx
\displaystyle = \sqrt{5}\int_0^{a} x^{1/2}\,dx
\displaystyle = \sqrt{5}\left[\frac{2}{3}x^{3/2}\right]_0^{a}
\displaystyle = \frac{2\sqrt{5}}{3}a^{3/2}
\displaystyle \text{Similarly, area } CABC \text{ is bounded by the circle}
\displaystyle \text{Area } CABC = \int_a^{6/\sqrt{5}} |y_2|\,dx
\displaystyle \text{Since } y_2 > 0,\ |y_2| = y_2
\displaystyle y_2 = \sqrt{\frac{36}{5} - x^2}
\displaystyle \text{Area } CABC = \int_a^{6/\sqrt{5}} \sqrt{\frac{36}{5} - x^2}\,dx
\displaystyle = \left[\frac{x}{2}\sqrt{\frac{36}{5} - x^2} + \frac{18}{5}\sin^{-1}\left(\frac{x\sqrt{5}}{6}\right)\right]_a^{6/\sqrt{5}}
\displaystyle = 0 + \frac{18}{5}\sin^{-1}(1) - \frac{a}{2}\sqrt{\frac{36}{5} - a^2} - \frac{18}{5}\sin^{-1}\left(\frac{a\sqrt{5}}{6}\right)
\displaystyle = \frac{9\pi}{5} - \frac{a}{2}\sqrt{\frac{36}{5} - a^2} - \frac{18}{5}\sin^{-1}\left(\frac{a\sqrt{5}}{6}\right)
\displaystyle \text{Hence required area } A = 2\left[\frac{2\sqrt{5}}{3}a^{3/2} + \frac{9\pi}{5} - \frac{a}{2}\sqrt{\frac{36}{5} - a^2} - \frac{18}{5}\sin^{-1}\left(\frac{a\sqrt{5}}{6}\right)\right]
\displaystyle A = \frac{4\sqrt{5}}{3}a^{3/2} + \frac{18\pi}{5} - a\sqrt{\frac{36}{5} - a^2} - \frac{36}{5}\sin^{-1}\left(\frac{a\sqrt{5}}{6}\right)
\displaystyle \text{where } a = \frac{-25 + \sqrt{1345}}{10}

\displaystyle \textbf{Question 15: }~\text{Draw a rough sketch and find the area of the region bounded by } \\ \text{the two parabolas }y^{2}=4x\text{ and }x^{2}=4y\text{ by using methods of integration.}
\displaystyle \text{Answer:} \displaystyle \text{To find the points of intersection between the two parabolas, substitute } x = \frac{y^2}{4} \text{ in } x^2 = 4y
\displaystyle \left(\frac{y^2}{4}\right)^2 = 4y
\displaystyle \Rightarrow \frac{y^4}{16} = 4y
\displaystyle \Rightarrow y^4 - 64y = 0
\displaystyle \Rightarrow y(y^3 - 64) = 0
\displaystyle \Rightarrow y = 0 \text{ or } y = 4
\displaystyle \Rightarrow x = \frac{y^2}{4} = 0 \text{ or } 4
\displaystyle \text{Therefore, the points of intersection are } A(4,4) \text{ and } C(0,0)
\displaystyle \text{Therefore, the area of the required region } ABCD
\displaystyle = \int_0^4 y_1\,dx - \int_0^4 y_2\,dx \text{ where } y_1 = 2\sqrt{x} \text{ and } y_2 = \frac{x^2}{4}
\displaystyle \text{Required Area}
\displaystyle = \int_0^4 (2\sqrt{x})\,dx - \int_0^4 \left(\frac{x^2}{4}\right)\,dx
\displaystyle = \left[2\cdot\frac{2x^{3/2}}{3} - \frac{x^3}{12}\right]_0^4
\displaystyle = \left(\frac{4(4)^{3/2}}{3} - \frac{(4)^3}{12}\right) - \left(\frac{4(0)^{3/2}}{3} - \frac{(0)^3}{12}\right)
\displaystyle = \frac{32}{3} - \frac{16}{3}
\displaystyle = \frac{16}{3} \text{ square units}

\displaystyle \textbf{Question 16: }~\text{Find the area included between the parabolas }y^{2}=4ax \\ \text{ and }x^{2}=4by.
\displaystyle \text{Answer:} \displaystyle \text{To find the point of intersection of the parabolas, substitute } y = \frac{x^2}{4b} \text{ in } y^2 = 4ax
\displaystyle \Rightarrow \left(\frac{x^2}{4b}\right)^2 = 4ax
\displaystyle \Rightarrow \frac{x^4}{16b^2} = 4ax
\displaystyle \Rightarrow x^4 - 64ab^2x = 0
\displaystyle \Rightarrow x(x^3 - 64ab^2) = 0
\displaystyle \Rightarrow x = 0 \text{ or } x = 4\sqrt[3]{ab^2}
\displaystyle \Rightarrow y = 0 \text{ or } y = \frac{(4\sqrt[3]{ab^2})^2}{4b}
\displaystyle \Rightarrow y = 4\sqrt[3]{a^2b}
\displaystyle \text{Therefore, the limits of integration are } x = 0 \text{ to } x = 4\sqrt[3]{ab^2}
\displaystyle \text{The required area } ABCD = \int_0^{4\sqrt[3]{ab^2}} (y_1 - y_2)\,dx
\displaystyle \text{where } y_1 = 2\sqrt{ax} \text{ and } y_2 = \frac{x^2}{4b}
\displaystyle \text{Required area } = \int_0^{4\sqrt[3]{ab^2}} \left(2\sqrt{ax} - \frac{x^2}{4b}\right) dx
\displaystyle = \left[\frac{4\sqrt{a}}{3}x^{3/2} - \frac{x^3}{12b}\right]_0^{4\sqrt[3]{ab^2}}
\displaystyle = \left[\frac{4\sqrt{a}}{3}(4\sqrt[3]{ab^2})^{3/2} - \frac{(4\sqrt[3]{ab^2})^3}{12b}\right]
\displaystyle = \left[\frac{4\sqrt{a}}{3}(8ab) - \frac{64a^3b^2}{12b}\right]
\displaystyle = \frac{32ab}{3} - \frac{16ab}{3}
\displaystyle = \frac{16ab}{3} \text{ square units}

\displaystyle \textbf{Question 17: }~\text{Prove that the area in the first quadrant enclosed by the } \\ x\text{-axis, the line }x=\sqrt{3}\,y\text{ and the circle }x^{2}+y^{2}=4\text{ is }\pi/3.
\displaystyle \text{Answer:} \displaystyle x^2 + y^2 = 4 \text{ represents a circle with centre } O(0,0) \text{ and radius } 2
\displaystyle \text{It cuts the } x\text{-axis at } A(2,0) \text{ and } A'(-2,0)
\displaystyle x = \sqrt{3}\,y \text{ represents a straight line passing through } O(0,0)
\displaystyle \text{Solving the two equations } x^2 + y^2 = 4 \text{ and } x = \sqrt{3}\,y
\displaystyle \Rightarrow (\sqrt{3}y)^2 + y^2 = 4
\displaystyle \Rightarrow 3y^2 + y^2 = 4
\displaystyle \Rightarrow 4y^2 = 4
\displaystyle \Rightarrow y = \pm 1
\displaystyle \Rightarrow x = \pm \sqrt{3}
\displaystyle \text{Thus } B(\sqrt{3},1) \text{ and } B'(-\sqrt{3},-1) \text{ are the points of intersection}
\displaystyle \text{Required shaded area } (OBQAO) = \text{area } (OBPO) + \text{area } (BAPB)
\displaystyle \text{Equation of line gives } y = \frac{x}{\sqrt{3}}
\displaystyle \text{Area } (OBPO) = \int_0^{\sqrt{3}} \frac{x}{\sqrt{3}}\,dx
\displaystyle \text{Area } (BAPB) = \int_{\sqrt{3}}^{2} \sqrt{4 - x^2}\,dx
\displaystyle \text{Required area } = \frac{1}{\sqrt{3}}\int_0^{\sqrt{3}} x\,dx + \int_{\sqrt{3}}^{2} \sqrt{4 - x^2}\,dx
\displaystyle = \frac{1}{\sqrt{3}}\left[\frac{x^2}{2}\right]_0^{\sqrt{3}} + \left[\frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\left(\frac{x}{2}\right)\right]_{\sqrt{3}}^{2}
\displaystyle = \frac{\sqrt{3}}{2} + \left[0 + 2\sin^{-1}(1) - \frac{\sqrt{3}}{2} - 2\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\right]
\displaystyle = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} + 2\left(\frac{\pi}{2} - \frac{\pi}{3}\right)
\displaystyle = \frac{\pi}{3} \text{ sq. units}
\displaystyle \text{Area bounded by the circle and straight line above the } x\text{-axis } = \frac{\pi}{3} \text{ sq. units}

\displaystyle \textbf{Question 18: }~\text{Find the area of the region bounded by }y=\sqrt{x},\  \\ x=2y+3\text{ in the first quadrant and }x\text{-axis.}
\displaystyle \text{Answer:} \displaystyle \text{The curve } y = \sqrt{x} \text{ or } y^2 = x \text{ represents a parabola opening towards the positive } x\text{-axis}
\displaystyle \text{The curve } x = 2y + 3 \text{ represents a straight line passing through } (3,0) \text{ and } (0,-\tfrac{3}{2})
\displaystyle \text{Solving } y^2 = x \text{ and } x = 2y + 3
\displaystyle \Rightarrow y^2 = 2y + 3
\displaystyle \Rightarrow y^2 - 2y - 3 = 0
\displaystyle \Rightarrow (y - 3)(y + 1) = 0
\displaystyle \Rightarrow y = 3 \text{ or } y = -1
\displaystyle \text{From the figure, the shaded region lies above the } x\text{-axis}
\displaystyle \text{Hence, } y \text{ varies from } 0 \text{ to } 3
\displaystyle \text{Required area } = \text{Area of the shaded region}
\displaystyle = \int_0^3 x_{\text{line}}\,dy - \int_0^3 x_{\text{parabola}}\,dy
\displaystyle = \int_0^3 (2y + 3)\,dy - \int_0^3 y^2\,dy
\displaystyle = \left[y^2 + 3y\right]_0^3 - \left[\frac{y^3}{3}\right]_0^3
\displaystyle = (9 + 9) - 9
\displaystyle = 9 \text{ square units}

\displaystyle \textbf{Question 19: }~\text{Find the area common to the circle }x^{2}+y^{2}=16a^{2} \\ \text{ and the parabola }y^{2}=6ax.
\displaystyle \text{Answer:} \displaystyle \text{Points of intersection of the parabola and the circle are obtained by solving}
\displaystyle x^2 + y^2 = 16a^2 \text{ and } y^2 = 6ax
\displaystyle \Rightarrow x^2 + 6ax = 16a^2
\displaystyle \Rightarrow x^2 + 6ax - 16a^2 = 0
\displaystyle \Rightarrow (x + 8a)(x - 2a) = 0
\displaystyle \Rightarrow x = 2a \text{ or } x = -8a
\displaystyle \text{Since the parabola lies for } x \ge 0,\ x = 2a \text{ is admissible}
\displaystyle \Rightarrow y^2 = 6a(2a) = 12a^2
\displaystyle \Rightarrow y = \pm 2\sqrt{3}a
\displaystyle \text{Thus } B(2a,2\sqrt{3}a) \text{ and } B'(2a,-2\sqrt{3}a) \text{ are points of intersection}
\displaystyle \text{Now, required area } = \text{area } (OBAB'O)
\displaystyle = 2(\text{area } OBAO)
\displaystyle = 2\{\text{area } OBDO + \text{area } DBAD\}
\displaystyle = 2\left[\int_0^{2a} \sqrt{6ax}\,dx + \int_{2a}^{4a} \sqrt{16a^2 - x^2}\,dx\right]
\displaystyle = 2\left[\sqrt{6a}\int_0^{2a} x^{1/2}\,dx + \int_{2a}^{4a} \sqrt{16a^2 - x^2}\,dx\right]
\displaystyle = 2\left[\sqrt{6a}\left(\frac{2}{3}x^{3/2}\right)_0^{2a} + \left(\frac{x}{2}\sqrt{16a^2 - x^2} + 8a^2\sin^{-1}\left(\frac{x}{4a}\right)\right)_{2a}^{4a}\right]
\displaystyle = 2\left[\frac{2\sqrt{6a}}{3}(2a)^{3/2} + \left(0 + 8a^2\sin^{-1}(1) - 2a\sqrt{12a^2} - 8a^2\sin^{-1}\left(\frac{1}{2}\right)\right)\right]
\displaystyle = 2\left[\frac{8\sqrt{3}a^2}{3} + 8a^2\left(\frac{\pi}{2} - \frac{\pi}{6}\right) - 2\sqrt{3}a^2\right]
\displaystyle = 2\left[\frac{8\sqrt{3} - 6\sqrt{3}}{3}a^2 + \frac{8\pi}{3}a^2\right]
\displaystyle = 2\left[\frac{2\sqrt{3}}{3}a^2 + \frac{8\pi}{3}a^2\right]
\displaystyle = \frac{4\sqrt{3}}{3}a^2 + \frac{16\pi}{3}a^2
\displaystyle = \frac{4a^2}{3}(4\pi + \sqrt{3})

\displaystyle \textbf{Question 20: }~\text{Find the area, lying above }x\text{-axis and included between the circle } \\ x^{2}+y^{2}=8x\text{ and the parabola }y^{2}=4x.
\displaystyle \text{Answer:} \displaystyle \text{The given equations are}
\displaystyle x^2 + y^2 = 8x \quad (1)
\displaystyle y^2 = 4x \quad (2)
\displaystyle \text{Clearly } x^2 + y^2 = 8x \text{ represents a circle with centre } (4,0) \text{ and radius } 4
\displaystyle \text{Also } y^2 = 4x \text{ is a parabola with vertex at the origin and axis along the } x\text{-axis opening in the positive direction}
\displaystyle \text{To find the points of intersection of the curves, solve } (1) \text{ and } (2)
\displaystyle \Rightarrow x^2 + 4x = 8x
\displaystyle \Rightarrow x^2 - 4x = 0
\displaystyle \Rightarrow x(x - 4) = 0
\displaystyle \Rightarrow x = 0 \text{ or } x = 4
\displaystyle \text{When } x = 4,\ y^2 = 16 \Rightarrow y = \pm 4
\displaystyle \text{Consider a vertical strip of width } dx
\displaystyle \text{Length of strip } = |y_2 - y_1|
\displaystyle \text{Required area } A = \int_0^4 |y_2 - y_1|\,dx
\displaystyle \text{Since } y_2 > y_1,\ |y_2 - y_1| = y_2 - y_1
\displaystyle y_2 = \sqrt{16 - (x - 4)^2},\quad y_1 = 2\sqrt{x}
\displaystyle A = \int_0^4 \left[\sqrt{16 - (x - 4)^2} - 2\sqrt{x}\right] dx
\displaystyle = \int_0^4 \sqrt{16 - (x - 4)^2}\,dx - \int_0^4 2\sqrt{x}\,dx
\displaystyle = \left[\frac{x - 4}{2}\sqrt{16 - (x - 4)^2} + 8\sin^{-1}\left(\frac{x - 4}{4}\right)\right]_0^4 - \left[\frac{4}{3}x^{3/2}\right]_0^4
\displaystyle = \left[0 - 8\sin^{-1}(-1)\right] - \frac{4}{3}(4)^{3/2}
\displaystyle = 8\left(\frac{\pi}{2}\right) - \frac{32}{3}
\displaystyle = 4\pi - \frac{32}{3}
\displaystyle \text{Hence the required area is } 4\pi - \frac{32}{3} \text{ square units}

\displaystyle \textbf{Question 21: }~\text{Find the area enclosed by the parabolas }y=5x^{2} \\ \text{ and }y=2x^{2}+9.
\displaystyle \text{Answer:} \displaystyle y = 5x^2 \text{ represents a parabola with vertex at } O(0,0) \text{ opening upwards and symmetric about the } y\text{-axis}
\displaystyle y = 2x^2 + 9 \text{ represents a wider parabola with vertex at } C(0,9)
\displaystyle \text{To find the points of intersection, solve the two equations}
\displaystyle 5x^2 = 2x^2 + 9
\displaystyle \Rightarrow 3x^2 = 9
\displaystyle \Rightarrow x = \pm \sqrt{3}
\displaystyle \Rightarrow y = 5(\sqrt{3})^2 = 15
\displaystyle \text{Thus } A(\sqrt{3},15) \text{ and } A'(-\sqrt{3},15) \text{ are the points of intersection}
\displaystyle \text{Shaded area } A'OA = 2 \times \text{area } (OCAO)
\displaystyle \text{Consider a vertical strip of length } |y_2 - y_1| \text{ and width } dx
\displaystyle \text{Area of approximating rectangle } = |y_2 - y_1|\,dx
\displaystyle \text{The strip moves from } x = 0 \text{ to } x = \sqrt{3}
\displaystyle \text{Area } (OCAO) = \int_0^{\sqrt{3}} |y_2 - y_1|\,dx
\displaystyle \text{Since } y_2 > y_1,\ |y_2 - y_1| = y_2 - y_1
\displaystyle y_2 = 2x^2 + 9,\quad y_1 = 5x^2
\displaystyle \text{Area } (OCAO) = \int_0^{\sqrt{3}} (2x^2 + 9 - 5x^2)\,dx
\displaystyle = \int_0^{\sqrt{3}} (9 - 3x^2)\,dx
\displaystyle = \left[9x - x^3\right]_0^{\sqrt{3}}
\displaystyle = 9\sqrt{3} - 3\sqrt{3}
\displaystyle = 6\sqrt{3} \text{ sq. units}
\displaystyle \text{Shaded area } B'A'AB = 2 \times 6\sqrt{3} = 12\sqrt{3} \text{ sq. units}
\displaystyle \text{Thus, the area enclosed by the two parabolas is } 12\sqrt{3} \text{ sq. units}

\displaystyle \textbf{Question 22: }~\text{Prove that the area common to the two parabolas }y=2x^{2} \\ \text{ and }y=x^{2}+4\text{ is }\frac{32}{3}\text{ sq. units.}
\displaystyle \text{Answer:} \displaystyle \text{We have two parabolas } y = 2x^2 \text{ and } y = x^2 + 4
\displaystyle \text{To find the points of intersection, solve the two equations}
\displaystyle 2x^2 = x^2 + 4
\displaystyle \Rightarrow x^2 = 4
\displaystyle \Rightarrow x = \pm 2
\displaystyle \Rightarrow y = 4
\displaystyle \text{Thus } A(2,4) \text{ and } A'(-2,4) \text{ are the points of intersection of the two parabolas}
\displaystyle \text{Shaded area } = 2 \times \text{area } (OCAO)
\displaystyle \text{Consider a vertical strip of length } |y_2 - y_1| \text{ and width } dx
\displaystyle \text{Area of approximating rectangle } = |y_2 - y_1|\,dx
\displaystyle \text{The strip moves from } x = 0 \text{ to } x = 2
\displaystyle \text{Area } (OCAO) = \int_0^2 |y_2 - y_1|\,dx
\displaystyle \text{Since } y_2 > y_1,\ |y_2 - y_1| = y_2 - y_1
\displaystyle y_2 = x^2 + 4,\quad y_1 = 2x^2
\displaystyle \text{Area } (OCAO) = \int_0^2 (x^2 + 4 - 2x^2)\,dx
\displaystyle = \int_0^2 (4 - x^2)\,dx
\displaystyle = \left[4x - \frac{x^3}{3}\right]_0^2
\displaystyle = 8 - \frac{8}{3}
\displaystyle = \frac{16}{3} \text{ sq. units}
\displaystyle \text{Shaded area } = 2 \times \frac{16}{3} = \frac{32}{3} \text{ sq. units}

\displaystyle \textbf{Question 23: }~\text{Using integration, find the area of the region bounded by } \\ \text{the triangle whose vertices are}
\displaystyle \text{(i): }(-1,2),\ (1,5)\text{ and }(3,4).
\displaystyle \text{Answer:} \displaystyle \text{Equation of } AB \text{ is } y - 5 = \frac{2 - 5}{-1 - 1}(x - 1)
\displaystyle \Rightarrow y - 5 = \frac{3}{2}(x - 1)
\displaystyle \Rightarrow y = \frac{3}{2}x + 5 - \frac{3}{2} = \frac{3x + 7}{2}
\displaystyle \text{Equation of } BC \text{ is } y - 4 = \frac{5 - 4}{1 - 3}(x - 3)
\displaystyle \Rightarrow y - 4 = -\frac{1}{2}(x - 3)
\displaystyle \Rightarrow y = -\frac{1}{2}x + 4 + \frac{3}{2} = \frac{-x + 11}{2}
\displaystyle \text{Equation of } CA \text{ is } y - 2 = \frac{4 - 2}{3 + 1}(x + 1)
\displaystyle \Rightarrow y - 2 = \frac{1}{2}(x + 1)
\displaystyle \Rightarrow y = \frac{1}{2}x + 2 + \frac{1}{2} = \frac{x + 5}{2}
\displaystyle \text{Required area } = \text{Area of region } ABEFA + \text{Area of region } BCDEB - \text{Area of region } ACDFA
\displaystyle = \int_{-1}^{1} y_{AB}\,dx + \int_{1}^{3} y_{BC}\,dx - \int_{-1}^{3} y_{CA}\,dx
\displaystyle = \int_{-1}^{1} \frac{3x + 7}{2}\,dx + \int_{1}^{3} \frac{-x + 11}{2}\,dx - \int_{-1}^{3} \frac{x + 5}{2}\,dx
\displaystyle = \frac{1}{2}\left[\frac{(3x + 7)^2}{3}\right]_{-1}^{1} + \frac{1}{2}\left[\frac{(-x + 11)^2}{-2}\right]_{1}^{3} - \frac{1}{2}\left[\frac{(x + 5)^2}{2}\right]_{-1}^{3}
\displaystyle = \frac{1}{12}(100 - 16) + \frac{1}{4}(64 - 100) - \frac{1}{4}(64 - 16)
\displaystyle = \frac{84}{12} + \frac{36}{4} - \frac{48}{4}
\displaystyle = 7 + 9 - 12
\displaystyle = 4 \text{ square units}

\displaystyle \text{(ii): }(-2,1),\ (0,4)\text{ and }(2,3).
\displaystyle \text{Answer:} \displaystyle \text{Let }A(-2,1),\ B(0,4),\ C(2,3).
\displaystyle \text{Equation of }AB\text{ is } y-4=\frac{4-1}{0-(-2)}(x-0)
\displaystyle \Rightarrow y-4=\frac{3}{2}x
\displaystyle \Rightarrow y=\frac{3}{2}x+4
\displaystyle \text{Equation of }BC\text{ is } y-4=\frac{3-4}{2-0}(x-0)
\displaystyle \Rightarrow y-4=-\frac{1}{2}x
\displaystyle \Rightarrow y=4-\frac{1}{2}x
\displaystyle \text{Equation of }CA\text{ is } y-1=\frac{3-1}{2-(-2)}(x+2)
\displaystyle \Rightarrow y-1=\frac{1}{2}(x+2)
\displaystyle \Rightarrow y=\frac{1}{2}x+2
\displaystyle \text{Required area }=\text{Area of region under }AB\text{ from }x=-2\text{ to }0+\text{Area of region under }BC\text{ from }x=0\text{ to }2-\text{Area of region under }CA\text{ from }x=-2\text{ to }2
\displaystyle =\int_{-2}^{0}y_{AB}\,dx+\int_{0}^{2}y_{BC}\,dx-\int_{-2}^{2}y_{CA}\,dx
\displaystyle =\int_{-2}^{0}\left(\frac{3}{2}x+4\right)\,dx+\int_{0}^{2}\left(4-\frac{1}{2}x\right)\,dx-\int_{-2}^{2}\left(\frac{1}{2}x+2\right)\,dx
\displaystyle =\left[\frac{3}{4}x^2+4x\right]_{-2}^{0}+\left[4x-\frac{x^2}{4}\right]_{0}^{2}-\left[\frac{x^2}{4}+2x\right]_{-2}^{2}
\displaystyle =\left[0-\left(\frac{3}{4}\cdot4+4(-2)\right)\right]+\left[\left(8-\frac{4}{4}\right)-0\right]-\left[\left(\frac{4}{4}+4\right)-\left(\frac{4}{4}-4\right)\right]
\displaystyle =\left[0-(3-8)\right]+\left[8-1\right]-\left[(1+4)-(1-4)\right]
\displaystyle =5+7-8
\displaystyle =4\text{ square units}

\displaystyle \textbf{Question 24: }~\text{Find the area of the region bounded by }y=\sqrt{x}\text{ and }y=x.
\displaystyle \text{Answer:} \displaystyle \text{The curve } y = \sqrt{x} \text{ or } y^2 = x \text{ represents a parabola opening towards the positive } x\text{-axis}
\displaystyle \text{The curve } y = x \text{ represents a straight line passing through the origin}
\displaystyle \text{Solving } y^2 = x \text{ and } y = x
\displaystyle \Rightarrow x^2 = x
\displaystyle \Rightarrow x^2 - x = 0
\displaystyle \Rightarrow x(x - 1) = 0
\displaystyle \Rightarrow x = 0 \text{ or } x = 1
\displaystyle \text{Thus, the curves intersect at } O(0,0) \text{ and } A(1,1)
\displaystyle \text{Required area } = \text{Area of the shaded region } OAO
\displaystyle = \int_0^1 y_{\text{parabola}}\,dx - \int_0^1 y_{\text{line}}\,dx
\displaystyle = \int_0^1 \sqrt{x}\,dx - \int_0^1 x\,dx
\displaystyle = \left[\frac{2}{3}x^{3/2}\right]_0^1 - \left[\frac{x^2}{2}\right]_0^1
\displaystyle = \frac{2}{3}(1 - 0) - \frac{1}{2}(1 - 0)
\displaystyle = \frac{2}{3} - \frac{1}{2}
\displaystyle = \frac{1}{6} \text{ square units}

\displaystyle \textbf{Question 25: }~\text{Find the area of the region in the first quadrant enclosed } \\ \text{by }x\text{-axis, the line }y=\sqrt{3}\,x\text{ and the circle }x^{2}+y^{2}=16.
\displaystyle \text{Answer:} \displaystyle x^2 + y^2 = 16 \text{ represents a circle with centre } O(0,0) \text{ and cuts the } x\text{-axis at } A(4,0)
\displaystyle y = \sqrt{3}\,x \text{ represents a straight line passing through } O(0,0)
\displaystyle \text{Point of intersection is obtained by solving the two equations}
\displaystyle x^2 + y^2 = 16 \text{ and } y = \sqrt{3}\,x
\displaystyle \Rightarrow x^2 + (\sqrt{3}x)^2 = 16
\displaystyle \Rightarrow 4x^2 = 16
\displaystyle \Rightarrow x = \pm 2
\displaystyle \Rightarrow y = \pm 2\sqrt{3}
\displaystyle \text{Thus } B(2,2\sqrt{3}) \text{ and } B'(-2,-2\sqrt{3}) \text{ are points of intersection of the circle and straight line}
\displaystyle \text{Shaded area } (OBQAO) = \text{area } (OBPO) + \text{area } (PBQAP)
\displaystyle \text{Equation of line gives } y = \sqrt{3}x
\displaystyle \text{Area } (OBPO) = \int_0^2 \sqrt{3}x\,dx
\displaystyle \text{Area } (PBQAP) = \int_2^4 \sqrt{16 - x^2}\,dx
\displaystyle \text{Required area} = \int_0^2 \sqrt{3}x\,dx + \int_2^4 \sqrt{16 - x^2}\,dx
\displaystyle = \sqrt{3}\left[\frac{x^2}{2}\right]_0^2 + \left[\frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right)\right]_2^4
\displaystyle = \sqrt{3}\cdot 2 + \left[0 + 8\sin^{-1}(1) - 2\sqrt{12} - 8\sin^{-1}\left(\frac{1}{2}\right)\right]
\displaystyle = 2\sqrt{3} + \left(4\pi - 4\sqrt{3} - \frac{4\pi}{3}\right)
\displaystyle = 2\sqrt{3} - 4\sqrt{3} + \frac{8\pi}{3}
\displaystyle = \frac{8\pi}{3} - 2\sqrt{3}
\displaystyle \text{Area bounded by the circle and straight line above the } x\text{-axis is } \frac{8\pi}{3} - 2\sqrt{3} \text{ sq. units}

\displaystyle \textbf{Question 26: }~\text{Find the area of the region bounded by the parabola } \\ y^{2}=2x+1\text{ and the line }x-y-1=0.
\displaystyle \text{Answer:} \displaystyle \text{We have the curves } y^2 = 2x + 1 \text{ and } x - y - 1 = 0
\displaystyle \text{To find the points of intersection of the curves, solve both equations}
\displaystyle \Rightarrow y^2 = 2(1 + y) + 1
\displaystyle \Rightarrow y^2 - 2y - 3 = 0
\displaystyle \Rightarrow (y - 3)(y + 1) = 0
\displaystyle \Rightarrow y = 3 \text{ or } y = -1
\displaystyle \Rightarrow x = 1 + y = 4 \text{ or } 0
\displaystyle \text{Consider a horizontal strip of length } |x_2 - x_1| \text{ and width } dy
\displaystyle \text{where } P(x_2,y) \text{ lies on the straight line and } Q(x_1,y) \text{ lies on the parabola}
\displaystyle \text{Area of approximating rectangle } = |x_2 - x_1|\,dy
\displaystyle \text{The strip moves from } y = -1 \text{ to } y = 3
\displaystyle \text{Required area } = \text{area } (OADO) = \int_{-1}^{3} |x_2 - x_1|\,dy
\displaystyle \text{Since } x_2 > x_1,\ |x_2 - x_1| = x_2 - x_1
\displaystyle x_2 = 1 + y,\quad x_1 = \frac{y^2 - 1}{2}
\displaystyle \text{Required area } = \int_{-1}^{3} \left[(1 + y) - \frac{1}{2}(y^2 - 1)\right] dy
\displaystyle = \int_{-1}^{3} \left(1 + y - \frac{1}{2}y^2 + \frac{1}{2}\right) dy
\displaystyle = \int_{-1}^{3} \left(\frac{3}{2} + y - \frac{1}{2}y^2\right) dy
\displaystyle = \left[\frac{3}{2}y + \frac{y^2}{2} - \frac{y^3}{6}\right]_{-1}^{3}
\displaystyle = \left(\frac{9}{2} + \frac{9}{2} - \frac{27}{6}\right) - \left(-\frac{3}{2} + \frac{1}{2} + \frac{1}{6}\right)
\displaystyle = \left(\frac{9}{2}\right) - \left(-\frac{5}{6}\right)
\displaystyle = \frac{16}{3} \text{ square units}
\displaystyle \text{Area enclosed by the line and the given parabola } = \frac{16}{3} \text{ square units}

\displaystyle \textbf{Question 27: }~\text{Find the area of the region bounded by the curves } \\ y=x-1\text{ and }(y-1)^{2}=4(x+1).
\displaystyle \text{Answer:} \displaystyle \text{We have the curves } y = x - 1 \text{ and } (y - 1)^2 = 4(x + 1)
\displaystyle \text{Substitute } y = x - 1 \text{ in } (y - 1)^2 = 4(x + 1)
\displaystyle \Rightarrow (x - 1 - 1)^2 = 4(x + 1)
\displaystyle \Rightarrow (x - 2)^2 = 4(x + 1)
\displaystyle \Rightarrow x^2 - 4x + 4 = 4x + 4
\displaystyle \Rightarrow x^2 - 8x = 0
\displaystyle \Rightarrow x(x - 8) = 0
\displaystyle \Rightarrow x = 0 \text{ or } x = 8
\displaystyle \Rightarrow y = -1 \text{ or } y = 7
\displaystyle \text{Consider a horizontal strip of length } |x_2 - x_1| \text{ and width } dy
\displaystyle \text{where } P(x_2,y) \text{ lies on the straight line and } Q(x_1,y) \text{ lies on the parabola}
\displaystyle \text{Area of approximating rectangle } = |x_2 - x_1|\,dy
\displaystyle \text{The strip moves from } y = -1 \text{ to } y = 7
\displaystyle \text{Required area } = \text{area } (OADO) = \int_{-1}^{7} |x_2 - x_1|\,dy
\displaystyle \text{Since } x_2 > x_1,\ |x_2 - x_1| = x_2 - x_1
\displaystyle x_2 = 1 + y,\quad x_1 = \frac{(y - 1)^2}{4} - 1
\displaystyle \text{Required area } = \int_{-1}^{7} \left[(1 + y) - \left(\frac{(y - 1)^2}{4} - 1\right)\right] dy
\displaystyle = \int_{-1}^{7} \left(2 + y - \frac{(y - 1)^2}{4}\right) dy
\displaystyle = \int_{-1}^{7} \left(2 + y - \frac{1}{4}(y - 1)^2\right) dy
\displaystyle = \left[2y + \frac{y^2}{2} - \frac{1}{12}(y - 1)^3\right]_{-1}^{7}
\displaystyle = \left(14 + \frac{49}{2} - \frac{1}{12}\cdot6^3\right) - \left(-2 + \frac{1}{2} + \frac{1}{12}\cdot2^3\right)
\displaystyle = \left(\frac{41}{2} - 18\right) - \left(-\frac{3}{2} + \frac{2}{3}\right)
\displaystyle = \frac{41}{2} - 18 + \frac{5}{6}
\displaystyle = \frac{64}{3} \text{ square units}
\displaystyle \text{Area enclosed by the line and the given parabola } = \frac{64}{3} \text{ square units}

\displaystyle \textbf{Question 28: }~\text{Find the area enclosed by the curve }y=-x^{2} \\ \text{ and the straight line }x+y+2=0.
\displaystyle \text{Answer:} \displaystyle \text{The curve } y = -x^2 \text{ represents a parabola opening towards the negative } y\text{-axis}
\displaystyle \text{The straight line } x + y + 2 = 0 \text{ passes through } (-2,0) \text{ and } (0,-2)
\displaystyle \text{Solving } y = -x^2 \text{ and } x + y + 2 = 0
\displaystyle \Rightarrow x - x^2 + 2 = 0
\displaystyle \Rightarrow x^2 - x - 2 = 0
\displaystyle \Rightarrow (x - 2)(x + 1) = 0
\displaystyle \Rightarrow x = 2 \text{ or } x = -1
\displaystyle \Rightarrow y = -(-1)^2 = -1 \text{ or } y = -(2)^2 = -4
\displaystyle \text{Thus, the curves intersect at } A(-1,-1) \text{ and } B(2,-4)
\displaystyle \text{Required area } = \text{Area of the shaded region } OABO
\displaystyle = \int_{-1}^{2} y_{\text{line}}\,dx - \int_{-1}^{2} y_{\text{parabola}}\,dx
\displaystyle = \int_{-1}^{2} (-(x + 2))\,dx - \int_{-1}^{2} (-x^2)\,dx
\displaystyle = -\int_{-1}^{2} (x + 2)\,dx + \int_{-1}^{2} x^2\,dx
\displaystyle = -\left[\frac{(x + 2)^2}{2}\right]_{-1}^{2} + \left[\frac{x^3}{3}\right]_{-1}^{2}
\displaystyle = -\frac{1}{2}(16 - 1) + \frac{1}{3}(8 - (-1))
\displaystyle = -\frac{15}{2} + 3
\displaystyle = \frac{9}{2} \text{ square units}

\displaystyle \textbf{Question 29: }~\text{Find the area bounded by the parabola }y=2-x^{2} \\ \text{ and the straight line }y+x=0.
\displaystyle \text{Answer:} \displaystyle \text{The graph of the parabola } y = 2 - x^2 \text{ and the line } x + y = 0 \text{ is as shown}
\displaystyle \text{To find the points of intersection between the parabola and the line, substitute } x = -y \text{ in } y = 2 - x^2
\displaystyle \Rightarrow y = 2 - (-y)^2
\displaystyle \Rightarrow y = 2 - y^2
\displaystyle \Rightarrow y^2 + y - 2 = 0
\displaystyle \Rightarrow (y - 1)(y + 2) = 0
\displaystyle \Rightarrow y = 1 \text{ or } y = -2
\displaystyle \Rightarrow x = -1 \text{ or } x = 2
\displaystyle \text{Therefore, the points of intersection are } A(-1,1) \text{ and } C(2,-2)
\displaystyle \text{The area of the required region } ABCD = \int_{-1}^{2} y_1\,dx - \int_{-1}^{2} y_2\,dx
\displaystyle \text{where } y_1 = 2 - x^2 \text{ and } y_2 = -x
\displaystyle \text{Required area}
\displaystyle = \int_{-1}^{2} \left[(2 - x^2) - (-x)\right] dx
\displaystyle = \int_{-1}^{2} (2 - x^2 + x)\,dx
\displaystyle = \left[2x - \frac{x^3}{3} + \frac{x^2}{2}\right]_{-1}^{2}
\displaystyle = \left[2(2) - \frac{(2)^3}{3} + \frac{(2)^2}{2}\right] - \left[2(-1) - \frac{(-1)^3}{3} + \frac{(-1)^2}{2}\right]
\displaystyle = \left(4 - \frac{8}{3} + 2\right) - \left(-2 + \frac{1}{3} + \frac{1}{2}\right)
\displaystyle = \frac{10}{3} + \frac{7}{6}
\displaystyle = \frac{9}{2} \text{ square units}

\displaystyle \textbf{Question 30: }~\text{Using the method of integration, find the area of the region bounded } \\ \text{by the following lines: }3x-y-3=0,\ 2x+y-12=0,\ x-2y-1=0.
\displaystyle \text{Answer:} \displaystyle \text{We have the straight lines}
\displaystyle 3x - y - 3 = 0 \quad (1)
\displaystyle 2x + y - 12 = 0 \quad (2)
\displaystyle x - 2y - 1 = 0 \quad (3)
\displaystyle \text{Solving } (1) \text{ and } (2)
\displaystyle \Rightarrow 3x - y - 3 = 0,\; 2x + y - 12 = 0
\displaystyle \Rightarrow 5x - 15 = 0
\displaystyle \Rightarrow x = 3
\displaystyle \Rightarrow y = 6
\displaystyle \text{Thus } B(3,6) \text{ is the point of intersection of } (1) \text{ and } (2)
\displaystyle \text{Solving } (1) \text{ and } (3)
\displaystyle \Rightarrow 3x - y - 3 = 0,\; x - 2y - 1 = 0
\displaystyle \Rightarrow 5x = 5
\displaystyle \Rightarrow x = 1
\displaystyle \Rightarrow y = 0
\displaystyle \text{Thus } A(1,0) \text{ is the point of intersection of } (1) \text{ and } (3)
\displaystyle \text{Solving } (2) \text{ and } (3)
\displaystyle \Rightarrow 2x + y - 12 = 0,\; x - 2y - 1 = 0
\displaystyle \Rightarrow 5x = 25
\displaystyle \Rightarrow x = 5
\displaystyle \Rightarrow y = 2
\displaystyle \text{Thus } C(5,2) \text{ is the point of intersection of } (2) \text{ and } (3)
\displaystyle \text{Now, the area of triangle } ABC
\displaystyle = \{\text{area under } (1) \text{ from } x=1 \text{ to } x=3\} + \{\text{area under } (2) \text{ from } x=3 \text{ to } x=5\}
\displaystyle \quad - \{\text{area under } (3) \text{ from } x=1 \text{ to } x=5\}
\displaystyle = \int_1^3 (3x - 3)\,dx + \int_3^5 (12 - 2x)\,dx - \int_1^5 \frac{x - 1}{2}\,dx
\displaystyle = \left[3\left(\frac{x^2}{2} - x\right)\right]_1^3 + \left[12x - x^2\right]_3^5 - \frac{1}{2}\left[\frac{x^2}{2} - x\right]_1^5
\displaystyle = 3\left(\frac{9}{2} - 3 - \frac{1}{2} + 1\right) + \left((60 - 25) - (36 - 9)\right) - \frac{1}{2}\left(\frac{25}{2} - 5 - \frac{1}{2} + 1\right)
\displaystyle = 3(2) + 8 - 4
\displaystyle = 10 \text{ square units}

\displaystyle \textbf{Question 31: }~\text{Sketch the region bounded by the curves }y=x^{2}+2,\  \\ y=x,\ x=0 \text{ and }x=1.\ \text{Also, find the area of this region.}
\displaystyle \text{Answer:} \displaystyle \text{We have the curves } y = x^2 + 2 \text{ and } y = x
\displaystyle \text{The parabola } y = x^2 + 2 \text{ and the line } y = x \text{ do not intersect}
\displaystyle x = 1 \text{ is a line parallel to the } y\text{-axis}
\displaystyle \text{Point of intersection of the parabola and } x = 1 \text{ is obtained by substituting } x = 1 \text{ in } y = x^2 + 2
\displaystyle \Rightarrow y = 1 + 2 = 3
\displaystyle \text{Point of intersection of the two straight lines is obtained by substituting } x = 1 \text{ in } y = x
\displaystyle \Rightarrow y = 1
\displaystyle \text{Consider a vertical strip of length } |y_2 - y_1| \text{ and width } dx
\displaystyle \text{where } P(x,y_2) \text{ lies on the parabola and } Q(x,y_1) \text{ lies on the line } y = x
\displaystyle \text{Shaded area } = \int_0^1 |y_2 - y_1|\,dx
\displaystyle \text{Since } y_2 > y_1,\ |y_2 - y_1| = y_2 - y_1
\displaystyle y_2 = x^2 + 2,\quad y_1 = x
\displaystyle \text{Shaded area } = \int_0^1 \left[(x^2 + 2) - x\right] dx
\displaystyle = \int_0^1 (x^2 - x + 2)\,dx
\displaystyle = \left[\frac{x^3}{3} - \frac{x^2}{2} + 2x\right]_0^1
\displaystyle = \frac{1}{3} - \frac{1}{2} + 2
\displaystyle = \frac{2 - 3 + 12}{6}
\displaystyle = \frac{11}{6} \text{ square units}
\displaystyle \text{Thus, the area enclosed by the parabola and the given two lines is } \frac{11}{6} \text{ square units}

\displaystyle \textbf{Question 32: }~\text{Find the area bounded by the curves }x=y^{2}\text{ and } \\ x=3-2y^{2}.
\displaystyle \text{Answer:} \displaystyle x = y^2 \text{ is a parabola opening towards the positive } x\text{-axis, with vertex at }  \\ O(0,0) \text{ and symmetric about the } x\text{-axis}
\displaystyle x = 3 - 2y^2 \text{ is a parabola opening towards the negative } x\text{-axis, with vertex at }  \\ A(3,0) \text{ and symmetric about the } x\text{-axis}
\displaystyle \text{To find the points of intersection of the two parabolas, solve}
\displaystyle x = y^2 \text{ and } x = 3 - 2y^2
\displaystyle \Rightarrow y^2 = 3 - 2y^2
\displaystyle \Rightarrow 3y^2 = 3
\displaystyle \Rightarrow y = \pm 1
\displaystyle \text{When } y = 1,\ x = 1 \text{ and when } y = -1,\ x = 1
\displaystyle \text{Thus } E(1,1) \text{ and } F(1,-1) \text{ are the points of intersection}
\displaystyle \text{The nature of the boundary changes at } E \text{ and } F
\displaystyle \text{Draw } EF \text{ parallel to the } y\text{-axis}
\displaystyle C(1,0) \text{ is the point of intersection of } EF \text{ with the } x\text{-axis}
\displaystyle \text{Since both curves are symmetric about the } x\text{-axis}
\displaystyle \text{Area of shaded region } OEAFO = 2 \times \text{Area } OEAO
\displaystyle = 2(\text{Area } OECO + \text{Area } CEAC) \quad (1)
\displaystyle \text{Area } OECO = \int_0^1 |y_1|\,dx \text{ where } P(x,y_1) \text{ lies on } x = y^2
\displaystyle \text{Since } y_1 > 0,\ |y_1| = y_1
\displaystyle \text{From } x = y^2,\ y_1 = \sqrt{x}
\displaystyle \text{Area } OECO = \int_0^1 \sqrt{x}\,dx
\displaystyle = \left[\frac{2}{3}x^{3/2}\right]_0^1
\displaystyle = \frac{2}{3} \text{ sq. units} \quad (2)
\displaystyle \text{Area } CEAC = \int_1^3 |y_2|\,dx \text{ where } Q(x,y_2) \text{ lies on } x = 3 - 2y^2
\displaystyle \text{Since } y_2 > 0,\ |y_2| = y_2
\displaystyle y_2 = \sqrt{\frac{3 - x}{2}}
\displaystyle \text{Area } CEAC = \int_1^3 \sqrt{\frac{3 - x}{2}}\,dx
\displaystyle = \frac{1}{\sqrt{2}}\int_1^3 \sqrt{3 - x}\,dx
\displaystyle = \frac{1}{\sqrt{2}}\left[\frac{2}{3}(3 - x)^{3/2}\right]_1^3
\displaystyle = \frac{1}{\sqrt{2}}\cdot\frac{2}{3}\left[0 - (2)^{3/2}\right]
\displaystyle = \frac{4}{3} \text{ sq. units} \quad (3)
\displaystyle \text{From } (1), (2) \text{ and } (3)
\displaystyle \text{Area of shaded region } OEAFO = 2\left(\frac{2}{3} + \frac{4}{3}\right)
\displaystyle = 2 \times 2
\displaystyle = 4 \text{ square units}

\displaystyle \textbf{Question 33: }~\text{Using integration, find the area of the triangle }ABC\text{ coordinates } \\ \text{of whose vertices are }A(4,1),\ B(6,6)\text{ and }C(8,4).
\displaystyle \text{Answer:} \displaystyle A(4,1),\,B(6,6)\ \text{and}\ C(8,4)\ \text{are three given points.}
\displaystyle \text{Equation of }AB\ \text{is given by}
\displaystyle y-1=\frac{6-1}{6-4}(x-4)
\displaystyle y-1=\frac{5}{2}(x-4)
\displaystyle y=\frac{5}{2}x-9\qquad (1)
\displaystyle \text{Equation of }BC\ \text{is given by}
\displaystyle y-6=\frac{4-6}{8-6}(x-6)
\displaystyle y-6=-(x-6)
\displaystyle y=-x+12\qquad (2)
\displaystyle \text{Equation of }CA\ \text{is given by}
\displaystyle y-4=\frac{1-4}{4-8}(x-8)
\displaystyle y-4=\frac{3}{4}(x-8)
\displaystyle y=\frac{3}{4}x-2\qquad (3)
\displaystyle \text{Required shaded area }(ABC)=\text{area}(ABD)+\text{area}(BDC)
\displaystyle \text{Consider a point }P(x,y_2)\ \text{on }AB\ \text{and }Q(x,y_1)\ \text{on }AD
\displaystyle \text{For a vertical strip of width }dx,\ \text{area}=|y_2-y_1|dx
\displaystyle \text{The strip moves from }x=4\ \text{to }x=6
\displaystyle \text{Area}(ABD)=\int_{4}^{6}|y_2-y_1|dx
\displaystyle =\int_{4}^{6}\left[\left(\frac{5}{2}x-9\right)-\left(\frac{3}{4}x-2\right)\right]dx
\displaystyle =\int_{4}^{6}\left(\frac{7}{4}x-7\right)dx
\displaystyle =\left[\frac{7}{4}\cdot\frac{x^2}{2}-7x\right]_{4}^{6}
\displaystyle =\frac{7}{8}(36-16)-7(6-4)
\displaystyle =\frac{35}{2}-14=\frac{7}{2}\ \text{sq. units}
\displaystyle \text{Consider a point }S(x,y_4)\ \text{on }BC\ \text{and }R(x,y_3)\ \text{on }DC
\displaystyle \text{The strip moves from }x=6\ \text{to }x=8
\displaystyle \text{Area}(BDC)=\int_{6}^{8}|y_4-y_3|dx
\displaystyle =\int_{6}^{8}\left[(-x+12)-\left(\frac{3}{4}x-2\right)\right]dx
\displaystyle =\int_{6}^{8}\left(-\frac{7}{4}x+14\right)dx
\displaystyle =\left[-\frac{7}{4}\cdot\frac{x^2}{2}+14x\right]_{6}^{8}
\displaystyle =-\frac{7}{8}(64-36)+14(8-6)
\displaystyle =-\frac{196}{8}+28=\frac{7}{2}\ \text{sq. units}
\displaystyle \text{Hence, shaded area }(ABC)=\frac{7}{2}+\frac{7}{2}=7\ \text{sq. units}

\displaystyle \textbf{Question 34: }~\text{Using integration find the area of the region: } \\ \{(x,y):\lvert x-1\rvert \le y \le \sqrt{5-x^{2}}\}.
\displaystyle \text{Answer:} \displaystyle |x-1|\le y\le \sqrt{5-x^2}
\displaystyle |x-1|=\sqrt{5-x^2}
\displaystyle x=2,-1
\displaystyle A=\int_{-1}^{2}\left(\sqrt{5-x^2}-|x-1|\right)dx
\displaystyle =\int_{-1}^{2}\sqrt{5-x^2}\,dx+\int_{-1}^{1}(x-1)\,dx+\int_{1}^{2}(1-x)\,dx
\displaystyle =\left[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}\sin^{-1}\!\left(\frac{x}{\sqrt{5}}\right)\right]_{-1}^{2}+\left[\frac{x^2}{2}-x\right]_{-1}^{1}+\left[x-\frac{x^2}{2}\right]_{1}^{2}
\displaystyle =\frac{5}{2}\left(\sin^{-1}\!\left(\frac{2}{\sqrt{5}}\right)+\sin^{-1}\!\left(\frac{1}{\sqrt{5}}\right)\right)+\frac{1}{2}

\displaystyle \textbf{Question 35: }~\text{Find the area of the region bounded by } \\ y=\lvert x-1\rvert\text{ and }y=1.
\displaystyle \text{Answer:} \displaystyle \text{We have,}
\displaystyle y=|x-1|
\displaystyle \Rightarrow y=\begin{cases}x-1,& x\ge 1\\[4pt]1-x,& x<1\end{cases}
\displaystyle y=x-1\ \text{is a straight line originating from }A(1,0)\ \text{and making an angle }45^\circ\ \text{with the }x\text{-axis}
\displaystyle y=1-x\ \text{is a straight line originating from }A(1,0)\ \text{and making an angle }135^\circ\ \text{with the }x\text{-axis}
\displaystyle y=1\ \text{is a straight line parallel to the }x\text{-axis and passing through }B(0,1)
\displaystyle \text{The point of intersection of the two lines with }y=1\ \text{is obtained by solving}
\displaystyle y=1\ \text{and}\ y=x-1
\displaystyle \Rightarrow 1=x-1
\displaystyle \Rightarrow x-2=0
\displaystyle \Rightarrow x=2
\displaystyle \Rightarrow C(2,1)\ \text{is the point of intersection of }y=x-1\ \text{and }y=1
\displaystyle y=1\ \text{and}\ y=1-x
\displaystyle \Rightarrow 1=1-x
\displaystyle \Rightarrow x=0
\displaystyle \Rightarrow B(0,1)\ \text{is the point of intersection of }y=1-x\ \text{and }y=1
\displaystyle \text{Since }y=|x-1|\ \text{changes character at }A(1,0),\ \text{consider point }P(1,1)\ \text{on }BC
\displaystyle \text{Required shaded area }(ABCA)=\text{area}(ABPA)+\text{area}(PCAP)
\displaystyle =\int_{0}^{1}\left[1-(1-x)\right]dx+\int_{1}^{2}\left[1-(x-1)\right]dx
\displaystyle =\int_{0}^{1}x\,dx+\int_{1}^{2}(2-x)\,dx
\displaystyle =\left[\frac{x^{2}}{2}\right]_{0}^{1}+\left[2x-\frac{x^{2}}{2}\right]_{1}^{2}
\displaystyle =\frac{1}{2}+\left[4-2-2+\frac{1}{2}\right]
\displaystyle =\frac{1}{2}+\frac{1}{2}=1\ \text{sq. unit}

\displaystyle \textbf{Question 36: }~\text{Find the area of the region in the first quadrant enclosed } \\ \text{by the }x\text{-axis, the line }y=x\text{ and the circle }x^{2}+y^{2}=32.
\displaystyle \text{Answer:} \displaystyle \text{We have,}
\displaystyle x^{2}+y^{2}=32\ \text{and}\ y=x
\displaystyle \text{The point of intersection of the circle and the straight line is obtained by solving the two equations}
\displaystyle x^{2}+x^{2}=32
\displaystyle \Rightarrow 2x^{2}=32
\displaystyle \Rightarrow x^{2}=16
\displaystyle \Rightarrow x=\pm 4
\displaystyle \Rightarrow y=\pm 4
\displaystyle \text{Thus }C(4,4)\ \text{and}\ C'(-4,-4)\ \text{are points of intersection of the circle and straight line}
\displaystyle \text{Required shaded area }(OCAPO)=\text{area}(OCPO)+\text{area}(PCAP)
\displaystyle =\int_{0}^{4}|y_1|\,dx+\int_{4}^{\sqrt{32}}|y_2|\,dx
\displaystyle =\int_{0}^{4}y_1\,dx+\int_{4}^{\sqrt{32}}y_2\,dx\ \ \{y_1>0\Rightarrow|y_1|=y_1,\ y_2>0\Rightarrow|y_2|=y_2\}
\displaystyle =\int_{0}^{4}x\,dx+\int_{4}^{\sqrt{32}}\sqrt{32-x^{2}}\,dx
\displaystyle =\left[\frac{x^{2}}{2}\right]_{0}^{4}+\left[\frac{1}{2}x\sqrt{32-x^{2}}+\frac{1}{2}\times 32\sin^{-1}\!\left(\frac{x}{\sqrt{32}}\right)\right]_{4}^{\sqrt{32}}
\displaystyle =8+8\pi-8-4\pi
\displaystyle =4\pi\ \text{sq. units}

\displaystyle \textbf{Question 37: }~\text{Find the area of the circle }x^{2}+y^{2}=16\text{ which is exterior } \\ \text{to the parabola }y^{2}=6x.
\displaystyle \text{Answer:} \displaystyle \text{Points of intersection of the parabola and the circle are obtained by solving the simultaneous equations}
\displaystyle x^{2}+y^{2}=16\ \text{and}\ y^{2}=6x
\displaystyle \Rightarrow x^{2}+6x=16
\displaystyle \Rightarrow x^{2}+6x-16=0
\displaystyle \Rightarrow (x+8)(x-2)=0
\displaystyle \Rightarrow x=2\ \text{or}\ x=-8,\ \text{which is not a possible solution}
\displaystyle \Rightarrow \text{When }x=2,\ y=\pm\sqrt{6\times2}=\pm\sqrt{12}=\pm2\sqrt{3}
\displaystyle \Rightarrow B(2,2\sqrt{3})\ \text{and}\ B'(2,-2\sqrt{3})\ \text{are points of intersection of the parabola and circle}
\displaystyle \text{Required area}=\text{Area}(OB'C'A'CBO)=\text{area of circle}-\text{area}(OBAB'O)
\displaystyle \text{Area of circle with radius }4=\pi\times4^{2}=16\pi
\displaystyle \text{Now,}
\displaystyle \text{Area}(OBAB'O)=2\times\text{area}(OBAO)
\displaystyle =2\{\text{area}(OBDO)+\text{area}(DBAD)\}
\displaystyle =2\left\{\int_{0}^{2}\sqrt{6x}\,dx+\int_{2}^{4}\sqrt{16-x^{2}}\,dx\right\}
\displaystyle =2\left\{\left[\sqrt{6}\frac{x^{3/2}}{3/2}\right]_{0}^{2}+\left[\frac{1}{2}x\sqrt{16-x^{2}}+\frac{1}{2}\times16\sin^{-1}\!\left(\frac{x}{4}\right)\right]_{2}^{4}\right\}
\displaystyle =2\left\{\left(\sqrt{6}\times\frac{2}{3}\times2\sqrt{2}\right)+0+8\sin^{-1}(1)-\sqrt{12}-8\sin^{-1}\!\left(\frac{1}{2}\right)\right\}
\displaystyle =2\left\{\frac{8\sqrt{3}}{3}+8\times\frac{\pi}{2}-2\sqrt{3}-8\times\frac{\pi}{6}\right\}
\displaystyle =2\left\{\frac{8\sqrt{3}-6\sqrt{3}}{3}+8\left(\frac{\pi}{2}-\frac{\pi}{6}\right)\right\}
\displaystyle =2\left\{\frac{2\sqrt{3}}{3}+8\times\frac{2\pi}{6}\right\}
\displaystyle =\frac{4\sqrt{3}}{3}+\frac{16\pi}{3}
\displaystyle \text{Shaded area}=16\pi-\left(\frac{4\sqrt{3}}{3}+\frac{16\pi}{3}\right)
\displaystyle =\frac{48\pi-16\pi}{3}-\frac{4\sqrt{3}}{3}
\displaystyle =\frac{32\pi}{3}-\frac{4\sqrt{3}}{3}
\displaystyle =\frac{4}{3}(8\pi-\sqrt{3})\ \text{sq. units}

\displaystyle \textbf{Question 38: }~\text{Find the area of the region enclosed by the parabola } \\ x^{2}=y\text{ and the line }y=x+2.
\displaystyle \text{Answer:} \displaystyle \text{The points of intersection }C\ \text{and }D\ \text{are obtained by solving the two equations}
\displaystyle x^{2}=x+2
\displaystyle \Rightarrow x^{2}-x-2=0
\displaystyle \Rightarrow (x-2)(x+1)=0
\displaystyle \Rightarrow x=2\ \text{or}\ x=-1
\displaystyle \Rightarrow y=2^{2}=4\ \text{or}\ y=(-1)^{2}=1
\displaystyle \text{Thus }C(2,4)\ \text{and }D(-1,1)\ \text{are points of intersection of the two curves}
\displaystyle \text{Consider a vertical strip of length }|y_2-y_1|\ \text{and width }dx\ \text{where }P(x,y_2)\ \text{lies on the straight line and }Q(x,y_1)\ \text{lies on the parabola}
\displaystyle \text{Area of approximating rectangle}=|y_2-y_1|dx,\ \text{and it moves from }x=-1\ \text{to }x=2
\displaystyle \text{Required area }=\text{area}(ODB C O)=\int_{-1}^{2}|y_2-y_1|dx
\displaystyle =\int_{-1}^{2}(y_2-y_1)\,dx\ \ \{y_2>y_1\Rightarrow|y_2-y_1|=y_2-y_1\}
\displaystyle =\int_{-1}^{2}\left[(x+2)-x^{2}\right]dx
\displaystyle =\int_{-1}^{2}(x+2-x^{2})dx
\displaystyle =\left[\frac{x^{2}}{2}+2x-\frac{x^{3}}{3}\right]_{-1}^{2}
\displaystyle =\frac{4}{2}+4-\frac{8}{3}-\left(\frac{1}{2}-2+\frac{1}{3}\right)
\displaystyle =\frac{9}{2}\ \text{sq. units}
\displaystyle \text{Area enclosed by the line and the given parabola }=\frac{9}{2}\ \text{sq. units}

\displaystyle \textbf{Question 39: }~\text{Make a sketch of the region }\{(x,y):0\le y\le x^{2}+3,\ 0\le y\le 2x+3,\ 0\le x\le 3\}\text{ and find its area using integration.}
\displaystyle \text{Answer:} \displaystyle R=\{(x,y):0\le y\le x^{2}+3,\ 0\le y\le 2x+3,\ 0\le x\le 3\}
\displaystyle R_{1}=\{(x,y):0\le y\le x^{2}+3\}
\displaystyle R_{2}=\{(x,y):0\le y\le 2x+3\}
\displaystyle R_{3}=\{(x,y):0\le x\le 3\}
\displaystyle \Rightarrow R=R_{1}\cap R_{2}\cap R_{3}
\displaystyle y=x^{2}+3\ \text{is an upward opening parabola with vertex }A(0,3)
\displaystyle \text{Thus }R_{1}\ \text{is the region above the }x\text{-axis and below the parabola}
\displaystyle y=2x+3\ \text{is a straight line passing through }A(0,3)\ \text{and cutting the }x\text{-axis at }(-\tfrac{3}{2},0)
\displaystyle \text{Hence }R_{2}\ \text{is the region above the }x\text{-axis and below the line}
\displaystyle x=3\ \text{is a straight line parallel to the }y\text{-axis and cuts the }x\text{-axis at }E(3,0)
\displaystyle \text{Hence }R_{3}\ \text{is the region above the }x\text{-axis and to the left of }x=3
\displaystyle \text{Point of intersection of the parabola and the line }y=2x+3\ \text{is obtained by solving}
\displaystyle y=x^{2}+3
\displaystyle y=2x+3
\displaystyle \Rightarrow x^{2}+3=2x+3
\displaystyle \Rightarrow x^{2}-2x=0
\displaystyle \Rightarrow x(x-2)=0
\displaystyle \Rightarrow x=0\ \text{or}\ x=2
\displaystyle \Rightarrow y=3\ \text{or}\ y=7
\displaystyle \Rightarrow A(0,3)\ \text{and}\ B(2,7)\ \text{are points of intersection}
\displaystyle \text{Also, }x=3\ \text{cuts the parabola at }C(3,12)
\displaystyle \text{and }x=3\ \text{cuts the line }y=2x+3\ \text{at }D(3,9)
\displaystyle \text{We require the area of the shaded region}
\displaystyle \text{Total shaded area}=\int_{0}^{2}(x^{2}+3)\,dx+\int_{2}^{3}(2x+3)\,dx
\displaystyle =\left[\frac{x^{3}}{3}+3x\right]_{0}^{2}+\left[x^{2}+3x\right]_{2}^{3}
\displaystyle =\left(\frac{8}{3}+6\right)+\left[(9+9)-(4+6)\right]
\displaystyle =\frac{8}{3}+6+8
\displaystyle =\frac{50}{3}\ \text{sq. units}

\displaystyle \textbf{Question 40: }~\text{Find the area of the region bounded by the curve }y=\sqrt{1-x^{2}},\  \\ \text{line }y=x\text{ and the positive }x\text{-axis.}
\displaystyle \text{Answer:} \displaystyle y=\sqrt{1-x^{2}}
\displaystyle \Rightarrow y^{2}=1-x^{2}
\displaystyle \Rightarrow x^{2}+y^{2}=1
\displaystyle \text{Hence,}
\displaystyle y=\sqrt{1-x^{2}}\ \text{represents the upper half of the circle }x^{2}+y^{2}=1\ \text{with centre }O(0,0)\ \text{and radius }1\ \text{unit}
\displaystyle y=x\ \text{represents the equation of a straight line passing through }O(0,0)
\displaystyle \text{Point of intersection is obtained by solving the two equations}
\displaystyle y=x
\displaystyle y=\sqrt{1-x^{2}}
\displaystyle \Rightarrow x=\sqrt{1-x^{2}}
\displaystyle \Rightarrow x^{2}=1-x^{2}
\displaystyle \Rightarrow 2x^{2}=1
\displaystyle \Rightarrow x=\pm\frac{1}{\sqrt{2}}
\displaystyle \Rightarrow y=\pm\frac{1}{\sqrt{2}}
\displaystyle D\!\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\ \text{and}\ D'\!\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)\ \text{are points of intersection of the circle and the straight line}
\displaystyle \text{And }D\!\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\ \text{lies on }y=\sqrt{1-x^{2}}\ \text{and }y=x
\displaystyle \text{Required area}=\text{shaded area }(OD AEO)
\displaystyle =\text{area}(ODEO)+\text{area}(EDAE)\qquad (1)
\displaystyle \text{Now, area}(ODEO)=\int_{0}^{\frac{1}{\sqrt{2}}}x\,dx
\displaystyle =\left[\frac{x^{2}}{2}\right]_{0}^{\frac{1}{\sqrt{2}}}
\displaystyle =\frac{1}{2}\left(\frac{1}{\sqrt{2}}\right)^{2}
\displaystyle =\frac{1}{4}\ \text{sq. units}\qquad (2)
\displaystyle \text{Area}(EDAE)=\int_{\frac{1}{\sqrt{2}}}^{1}\sqrt{1-x^{2}}\,dx
\displaystyle =\left[\frac{1}{2}x\sqrt{1-x^{2}}+\frac{1}{2}\sin^{-1}(x)\right]_{\frac{1}{\sqrt{2}}}^{1}
\displaystyle =0+\frac{1}{2}\sin^{-1}(1)-\frac{1}{2}\cdot\frac{1}{\sqrt{2}}\cdot\sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^{2}}-\frac{1}{2}\sin^{-1}\!\left(\frac{1}{\sqrt{2}}\right)
\displaystyle =\frac{1}{2}\cdot\frac{\pi}{2}-\frac{1}{2}\cdot\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}-\frac{1}{2}\cdot\frac{\pi}{4}
\displaystyle =\frac{\pi}{4}-\frac{1}{8}-\frac{\pi}{8}
\displaystyle =\frac{\pi}{8}-\frac{1}{8}\ \text{sq. units}\qquad (3)
\displaystyle \text{From }(1),(2)\ \text{and }(3),\ \text{we get}
\displaystyle \text{Area}(OD AEO)=\frac{1}{4}+\frac{\pi}{8}-\frac{1}{4}=\frac{\pi}{8}\ \text{sq. units}

\displaystyle \textbf{Question 41: }~\text{Find the area bounded by the lines }y=4x+5,\ y=5-x \\ \text{ and }4y=x+5.
\displaystyle \text{Answer:} \displaystyle \text{We have,}
\displaystyle y=4x+5\qquad (1)
\displaystyle y=5-x\qquad (2)
\displaystyle 4y=x+5\qquad (3)
\displaystyle \text{All the three equations represent straight lines}
\displaystyle \text{The points of intersection are obtained by solving simultaneous equations}
\displaystyle \text{From }(1)\ \text{and }(2)
\displaystyle 4x+5=5-x
\displaystyle \Rightarrow 5x=0
\displaystyle \Rightarrow x=0
\displaystyle \Rightarrow y=5
\displaystyle \text{Thus }A(0,5)\ \text{is the point of intersection of }(1)\ \text{and }(2)
\displaystyle \text{From }(2)\ \text{and }(3)
\displaystyle 4(5-x)=x+5
\displaystyle \Rightarrow 5x=15
\displaystyle \Rightarrow x=3
\displaystyle \Rightarrow y=2
\displaystyle \text{Thus }B(3,2)\ \text{is the point of intersection of }(2)\ \text{and }(3)
\displaystyle \text{From }(1)\ \text{and }(3)
\displaystyle 4(4x+5)=x+5
\displaystyle \Rightarrow 15x=-15
\displaystyle \Rightarrow x=-1
\displaystyle \Rightarrow y=1
\displaystyle \text{Thus }C(-1,1)\ \text{is the point of intersection of }(1)\ \text{and }(3)
\displaystyle \text{Area}(ABC)=\text{area}(ABP)+\text{area}(PAB)
\displaystyle =\int_{-1}^{0}\left[(4x+5)-\frac{x+5}{4}\right]dx+\int_{0}^{3}\left[(5-x)-\frac{x+5}{4}\right]dx
\displaystyle =\int_{-1}^{0}\left(\frac{15}{4}x+\frac{15}{4}\right)dx+\int_{0}^{3}\left(\frac{15}{4}-\frac{5}{4}x\right)dx
\displaystyle =\frac{15}{4}\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}+\frac{5}{4}\left[3x-\frac{x^{2}}{2}\right]_{0}^{3}
\displaystyle =\frac{15}{4}\left(-\frac{1}{2}+1\right)+\frac{5}{4}\left(9-\frac{9}{2}\right)
\displaystyle =\frac{15}{8}+\frac{45}{8}
\displaystyle =\frac{60}{8}
\displaystyle =\frac{15}{2}\ \text{sq. units}

\displaystyle \textbf{Question 42: }~\text{Find the area of the region enclosed between the two curves } \\ x^{2}+y^{2}=9\text{ and }(x-3)^{2}+y^{2}=9.
\displaystyle \text{Answer:} \displaystyle \text{Let the two curves be named as }y_1\ \text{and }y_2\ \text{where}
\displaystyle y_1:(x-3)^2+y^2=9\qquad (1)
\displaystyle y_2:x^2+y^2=9\qquad (2)
\displaystyle \text{The curve }x^2+y^2=9\ \text{represents a circle with centre }(0,0)\ \text{and radius }3
\displaystyle \text{The curve }(x-3)^2+y^2=9\ \text{represents a circle with centre }(3,0)\ \text{and radius }3
\displaystyle \text{To find the points of intersection of the two curves, equate }(1)\ \text{and }(2)
\displaystyle x^2=(x-3)^2
\displaystyle \Rightarrow x^2=x^2-6x+9
\displaystyle \Rightarrow 6x=9
\displaystyle \Rightarrow x=\frac{3}{2}
\displaystyle \Rightarrow y^2=9-\left(\frac{3}{2}\right)^2=\frac{27}{4}
\displaystyle \Rightarrow y=\pm\frac{3\sqrt{3}}{2}
\displaystyle \text{Therefore, the points of intersection are }\left(\frac{3}{2},\frac{3\sqrt{3}}{2}\right)\ \text{and }\left(\frac{3}{2},-\frac{3\sqrt{3}}{2}\right)
\displaystyle \text{Now, the required area }(OABO)=2\,[\text{area}(OACO)+\text{area}(CABC)]
\displaystyle \text{Here,}
\displaystyle \text{area}(OACO)=\int_{0}^{\frac{3}{2}}y_1\,dx
\displaystyle =\int_{0}^{\frac{3}{2}}\sqrt{9-(x-3)^2}\,dx
\displaystyle \text{And}
\displaystyle \text{area}(CABC)=\int_{\frac{3}{2}}^{3}|y_2|\,dx
\displaystyle =\int_{\frac{3}{2}}^{3}\sqrt{9-x^2}\,dx
\displaystyle \text{Thus, the required area is given by}
\displaystyle A=2\left[\int_{0}^{\frac{3}{2}}\sqrt{9-(x-3)^2}\,dx+\int_{\frac{3}{2}}^{3}\sqrt{9-x^2}\,dx\right]
\displaystyle =2\left[\left(\frac{x-3}{2}\sqrt{9-(x-3)^2}+\frac{9}{2}\sin^{-1}\!\left(\frac{x-3}{3}\right)\right)_{0}^{\frac{3}{2}}+\left(\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}\sin^{-1}\!\left(\frac{x}{3}\right)\right)_{\frac{3}{2}}^{3}\right]
\displaystyle =2\left[\left(0-\left(-\frac{3}{2}\cdot\frac{3\sqrt{3}}{2}+\frac{9}{2}\sin^{-1}(-1)\right)\right)+\left(0-\left(\frac{3}{4}\cdot\frac{3\sqrt{3}}{2}+\frac{9}{2}\sin^{-1}\!\left(\frac{1}{2}\right)\right)\right)\right]
\displaystyle =2\left[\frac{9\sqrt{3}}{8}+\frac{9\pi}{4}-\frac{9\sqrt{3}}{8}-\frac{3\pi}{4}\right]
\displaystyle =2\left[\frac{3\pi}{2}-\frac{9\sqrt{3}}{4}\right]
\displaystyle =6\pi-\frac{9\sqrt{3}}{2}\ \text{sq. units}

\displaystyle \textbf{Question 43: }~\text{Find the area of the region }\{(x,y):x^{2}+y^{2}\le 4,\ x+y\ge 2\}.
\displaystyle \text{Answer:} \displaystyle R=\{(x,y):x^{2}+y^{2}\le4,\ x+y\ge2\}
\displaystyle R_{1}=\{(x,y):x^{2}+y^{2}\le4\}
\displaystyle R_{2}=\{(x,y):x+y\ge2\}
\displaystyle \Rightarrow R=R_{1}\cap R_{2}
\displaystyle \text{The region }R_{1}\ \text{represents the interior of the circle }x^{2}+y^{2}=4\ \text{with centre }(0,0)\ \text{and radius }2
\displaystyle \text{The region }R_{2}\ \text{lies above the line }x+y=2
\displaystyle \text{The line }x+y=2\ \text{and the circle }x^{2}+y^{2}=4\ \text{intersect at }(2,0)\ \text{and }(0,2)
\displaystyle \text{Here, the length of the shaded region is }|y_{2}-y_{1}|
\displaystyle \text{where }y_{2}\ \text{is }y\ \text{for the circle }x^{2}+y^{2}=4\ \text{and }y_{1}\ \text{is }y\ \text{for the line }x+y=2
\displaystyle \text{Since }y_{2}>y_{1},\ |y_{2}-y_{1}|=y_{2}-y_{1},\ \text{and the width is }dx
\displaystyle \text{Therefore, the area }A=\int_{0}^{2}(y_{2}-y_{1})\,dx
\displaystyle =\int_{0}^{2}\left[\sqrt{4-x^{2}}-(2-x)\right]dx
\displaystyle =\left[\frac{1}{2}x\sqrt{4-x^{2}}+2\sin^{-1}\!\left(\frac{x}{2}\right)\right]_{0}^{2}-\left[2x-\frac{x^{2}}{2}\right]_{0}^{2}
\displaystyle =\left[0+2\sin^{-1}(1)-0-0\right]-\left[4-2-0+0\right]
\displaystyle =2\sin^{-1}(1)-2
\displaystyle =2\times\frac{\pi}{2}-2
\displaystyle =\pi-2\ \text{sq. units}

\displaystyle \textbf{Question 44: }~\text{Using integration, find the area of the following region: } \\ \{(x,y):\frac{x^{2}}{9}+\frac{y^{2}}{4}\le 1\le \frac{x}{3}+\frac{y}{2}\}.
\displaystyle \text{Answer:} \displaystyle \text{Let } R=\left\{(x,y):\frac{x^{2}}{9}+\frac{y^{2}}{4}\le 1\le \frac{x}{3}+\frac{y}{2}\right\}
\displaystyle R_{1}=\left\{(x,y):\frac{x^{2}}{9}+\frac{y^{2}}{4}\le 1\right\}
\displaystyle R_{2}=\left\{(x,y):1\le \frac{x}{3}+\frac{y}{2}\right\}
\displaystyle \therefore R=R_{1}\cap R_{2}
\displaystyle \frac{x^{2}}{9}+\frac{y^{2}}{4}=1 \text{ represents an ellipse with centre } O(0,0)
\displaystyle \text{It cuts the coordinate axes at } A(3,0),A'(-3,0),B(0,2)\text{ and }B'(0,-2)
\displaystyle \text{Hence } R_{1} \text{ is the region interior to the ellipse}
\displaystyle \frac{x}{3}+\frac{y}{2}=1
\displaystyle \Rightarrow 2x+3y=6 \text{ represents a straight line}
\displaystyle \text{The line cuts the axes at } A(3,0) \text{ and } B(0,2)
\displaystyle \text{Hence } R_{2} \text{ is the region above the line}
\displaystyle \text{Thus } A(3,0) \text{ and } B(0,2) \text{ are points of intersection}
\displaystyle \text{Area of the shaded region } A
\displaystyle A=\int_{0}^{3}\left[y_{2}-y_{1}\right]dx
\displaystyle A=\int_{0}^{3}\left[\sqrt{4\left(1-\frac{x^{2}}{9}\right)}-2\left(1-\frac{x}{3}\right)\right]dx
\displaystyle =\int_{0}^{3}\left[\frac{\sqrt{36-4x^{2}}}{3}-\frac{6-2x}{3}\right]dx
\displaystyle =\frac{1}{3}\int_{0}^{3}\left[2\sqrt{9-x^{2}}-(6-2x)\right]dx
\displaystyle =\frac{1}{3}\left[2\left\{\frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\sin^{-1}\!\left(\frac{x}{3}\right)\right\}-(6x-x^{2})\right]_{0}^{3}
\displaystyle =\frac{1}{3}\left[0+9\sin^{-1}(1)-18+9\right]
\displaystyle =\frac{1}{3}\left(\frac{9\pi}{2}-9\right)
\displaystyle =\left(\frac{3\pi}{2}-3\right)\text{ square units}

\displaystyle \textbf{Question 45: }~\text{Using integration find the area of the region bounded } \\ \text{by the curve }y=\sqrt{4-x^{2}},\ x^{2}+y^{2}-4x=0\text{ and the }x\text{-axis.}
\displaystyle \text{Answer:} \displaystyle \text{The given curves are } y=\sqrt{4-x^{2}} \text{ and } x^{2}+y^{2}-4x=0
\displaystyle y=\sqrt{4-x^{2}} \Rightarrow x^{2}+y^{2}=4 \qquad (1)
\displaystyle \text{Equation (1) represents a circle with centre } O(0,0) \text{ and radius } 2
\displaystyle x^{2}+y^{2}-4x=0 \Rightarrow (x-2)^{2}+y^{2}=4 \qquad (2)
\displaystyle \text{Equation (2) represents a circle with centre } B(2,0) \text{ and radius } 2
\displaystyle \text{To find the points of intersection, solve (1) and (2)}
\displaystyle (x-2)^{2}+y^{2}=x^{2}+y^{2}
\displaystyle x^{2}-4x+4=x^{2}
\displaystyle x=1
\displaystyle \text{Substituting } x=1 \text{ in } x^{2}+y^{2}=4
\displaystyle y^{2}=3 \Rightarrow y=\pm\sqrt{3}
\displaystyle \text{Thus, the points of intersection are } A(1,\sqrt{3}) \text{ and } C(1,-\sqrt{3})
\displaystyle \text{Required area = Area of the shaded region } OABO
\displaystyle =\int_{0}^{1}\sqrt{4-(x-2)^{2}}\,dx+\int_{1}^{2}\sqrt{4-x^{2}}\,dx
\displaystyle =\left[\frac{(x-2)}{2}\sqrt{4-(x-2)^{2}}+2\sin^{-1}\!\left(\frac{x-2}{2}\right)\right]_{0}^{1}
\displaystyle +\left[\frac{x}{2}\sqrt{4-x^{2}}+2\sin^{-1}\!\left(\frac{x}{2}\right)\right]_{1}^{2}
\displaystyle =\left[-\frac{\sqrt{3}}{2}+2\sin^{-1}\!\left(-\frac{1}{2}\right)\right]-\left[0+2\sin^{-1}(-1)\right]
\displaystyle +\left[0+2\sin^{-1}(1)\right]-\left[\frac{\sqrt{3}}{2}+2\sin^{-1}\!\left(\frac{1}{2}\right)\right]
\displaystyle =-\sqrt{3}+2\pi-\frac{2\pi}{3}
\displaystyle =\frac{4\pi}{3}-\sqrt{3}\ \text{square units}

\displaystyle \textbf{Question 46: }~\text{Find the area enclosed by the curves } \\ y=\lvert x-1\rvert\text{ and }y=-\lvert x-1\rvert+1.
\displaystyle \text{Answer:} \displaystyle \text{The given curves are } y=|x-1| \text{ .....(1)}
\displaystyle y=-|x-1|+1 \text{ .....(2)}
\displaystyle \text{Clearly, } y=|x-1| \text{ cuts the x-axis at } (1,0) \text{ and the y-axis at } (0,1).
\displaystyle \text{Also, } y=-|x-1|+1 \text{ cuts both the axes at } (0,0) \text{ and the x-axis at } (2,0).
\displaystyle y=|x-1|=\begin{cases}x-1,& x\ge 1\\1-x,& x<1\end{cases}
\displaystyle y=-|x-1|+1=\begin{cases}2-x,& x\ge 1\\x,& x<1\end{cases}
\displaystyle \text{Solving for } x<1 \text{ : } y=1-x \text{ and } y=x
\displaystyle 1-x=x
\displaystyle x=\frac12,\; y=\frac12
\displaystyle \text{Solving for } x\ge 1 \text{ : } y=x-1 \text{ and } y=2-x
\displaystyle x-1=2-x
\displaystyle x=\frac32,\; y=\frac12
\displaystyle \text{Thus, the points of intersection are } \left(\frac12,\frac12\right) \text{ and } \left(\frac32,\frac12\right).
\displaystyle \text{Required area } A=\text{Area(ABFA)}+\text{Area(BCFB)}
\displaystyle \text{Area(ABFA)}=\int_{1/2}^{1}\left[x-(1-x)\right]dx
\displaystyle =\int_{1/2}^{1}(2x-1)\,dx
\displaystyle =\left[x^{2}-x\right]_{1/2}^{1}
\displaystyle =\frac14
\displaystyle \text{Area(BCFB)}=\int_{1}^{3/2}\left[(2-x)-(x-1)\right]dx
\displaystyle =\int_{1}^{3/2}(3-2x)\,dx
\displaystyle =\left[3x-x^{2}\right]_{1}^{3/2}
\displaystyle =\frac14
\displaystyle A=\frac14+\frac14=\frac12
\displaystyle \text{Hence, the required area is } \frac12 \text{ square units.}

\displaystyle \textbf{Question 47: }~\text{Find the area enclosed by the curves }3x^{2}+5y=32 \\ \text{ and }y=\lvert x-2\rvert.
\displaystyle \text{Answer:} \displaystyle 3x^{2}+5y=32 \text{ represents a downward opening parabola, symmetrical about the y-axis.}
\displaystyle \text{Vertex of the parabola is } D\left(0,\frac{32}{5}\right).
\displaystyle \text{It cuts the x-axis at } B\left(\sqrt{\frac{32}{3}},0\right) \text{ and } B'\left(-\sqrt{\frac{32}{3}},0\right).
\displaystyle y=|x-2|
\displaystyle y=\begin{cases}x-2,& x\ge 2\\2-x,& x<2\end{cases}
\displaystyle y=2-x,\; x<2 \text{ represents a straight line cutting the x-axis at } A(2,0) \text{ and the parabola at } C(3,1).
\displaystyle \text{Also, } y=2-x,\; x<2 \text{ cuts the parabola at } (-2,4).
\displaystyle \text{Required shaded area } AEYCA
\displaystyle A=\int_{-2}^{2}\left(\frac{32-3x^{2}}{5}-(2-x)\right)dx+\int_{2}^{3}\left(\frac{32-3x^{2}}{5}-(x-2)\right)dx
\displaystyle =\int_{-2}^{3}\frac{32-3x^{2}}{5}dx-\int_{-2}^{2}(2-x)dx-\int_{2}^{3}(x-2)dx
\displaystyle =\frac15\left[32x-x^{3}\right]_{-2}^{3}-\left[2x-\frac{x^{2}}{2}\right]_{-2}^{2}-\left[\frac{x^{2}}{2}-2x\right]_{2}^{3}
\displaystyle =\frac15\left[(96-27)-(-64+8)\right]-\left[(4-2)-(-4-2)\right]-\left[\left(\frac92-6\right)-\left(2-4\right)\right]
\displaystyle =\frac15(125)-8-\left(\frac92-4\right)
\displaystyle =25-8-\frac92+4
\displaystyle =21-\frac92
\displaystyle =\frac{42-9}{2}
\displaystyle =\frac{33}{2}\text{ square units.}

\displaystyle \textbf{Question 48: }~\text{Find the area enclosed by the parabolas }y=4x-x^{2} \\ \text{ and }y=x^{2}-x.
\displaystyle \text{Answer:} \displaystyle \text{We have,}
\displaystyle y = 4x - x^{2} \text{ and } y = x^{2} - x
\displaystyle \text{The points of intersection of the two curves are obtained by solving the simultaneous equations}
\displaystyle x^{2} - x = 4x - x^{2}
\displaystyle 2x^{2} - 5x = 0
\displaystyle x(2x - 5) = 0
\displaystyle x = 0 \text{ or } x = \frac{5}{2}
\displaystyle y = 0 \text{ or } y = \frac{15}{4}
\displaystyle \text{Thus } O(0,0) \text{ and } D\left(\frac{5}{2},\frac{15}{4}\right) \text{ are the points of intersection of the two parabolas}
\displaystyle \text{In the shaded region CBDC, consider } P(x,y_{2}) \text{ on } y = 4x - x^{2} \text{ and } Q(x,y_{1}) \text{ on } y = x^{2} - x
\displaystyle \text{Area }(OBDCO) = \text{Area }(OBCO) + \text{Area }(CBDC)
\displaystyle = \int_{0}^{1} |y|\,dx + \int_{1}^{\frac{5}{2}} |y_{2}-y_{1}|\,dx
\displaystyle = \int_{0}^{1} y\,dx + \int_{1}^{\frac{5}{2}} (y_{2}-y_{1})\,dx
\displaystyle = \int_{0}^{1} (4x - x^{2})\,dx + \int_{1}^{\frac{5}{2}} \big[(4x - x^{2}) - (x^{2} - x)\big]\,dx
\displaystyle = \left[\frac{4x^{2}}{2} - \frac{x^{3}}{3}\right]_{0}^{1} + \int_{1}^{\frac{5}{2}} (5x - 2x^{2})\,dx
\displaystyle = \left[2x^{2} - \frac{x^{3}}{3}\right]_{0}^{1} + \left[\frac{5x^{2}}{2} - \frac{2x^{3}}{3}\right]_{1}^{\frac{5}{2}}
\displaystyle = \left(2 - \frac{1}{3}\right) + \left[\frac{5}{2}\left(\frac{5}{2}\right)^{2} - \frac{2}{3}\left(\frac{5}{2}\right)^{3} - \left(\frac{5}{2} - \frac{2}{3}\right)\right]
\displaystyle = \frac{5}{3} + \left[\frac{125}{8} - \frac{125}{12} - \frac{11}{6}\right]
\displaystyle = \frac{5}{3} + \frac{111}{24}
\displaystyle = \frac{121}{24} \text{ sq units}
\displaystyle \text{Now consider the region } OCV'O
\displaystyle \text{Area }(OCV'O) = \int_{0}^{1} |y|\,dx
\displaystyle = \int_{0}^{1} -(x^{2}-x)\,dx
\displaystyle = \int_{0}^{1} (x - x^{2})\,dx
\displaystyle = \left[\frac{x^{2}}{2} - \frac{x^{3}}{3}\right]_{0}^{1}
\displaystyle = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \text{ sq units}
\displaystyle \text{Hence, the required shaded area}
\displaystyle = \frac{121}{24} + \frac{1}{6}
\displaystyle = \frac{125}{24} \text{ sq units}

\displaystyle \textbf{Question 49: }~\text{In what ratio does the }x\text{-axis divide the area of the region } \\ \text{bounded by the parabolas }y=4x-x^{2}\text{ and }y=x^{2}-x?
\displaystyle \text{Answer:} \displaystyle \text{We have, }
\displaystyle y=4x-x^2 \text{ and } y=x^2-x
\displaystyle \text{The points of intersection of the two curves are obtained by solving the simultaneous equations}
\displaystyle x^2-x=4x-x^2
\displaystyle 2x^2-5x=0
\displaystyle x=0 \text{ or } x=\frac{5}{2}
\displaystyle y=0 \text{ or } y=\frac{15}{4}
\displaystyle \therefore O(0,0) \text{ and } D\!\left(\frac{5}{2},\frac{15}{4}\right) \text{ are points of intersection of the two parabolas}
\displaystyle \text{In the shaded region CBDC, consider } P(x,y_2) \text{ on } y=4x-x^2 \text{ and } Q(x,y_1) \text{ on } y=x^2-x
\displaystyle \text{We need to find the ratio of areas } (OBDCO) \text{ and } (OCV'O)
\displaystyle \text{Area }(OBDCO)=\text{area }(OBCO)+\text{area }(CBDC)
\displaystyle =\int_{0}^{1}|y|\,dx+\int_{1}^{\frac{5}{2}}|y_2-y_1|\,dx
\displaystyle =\int_{0}^{1}y\,dx+\int_{1}^{\frac{5}{2}}(y_2-y_1)\,dx \quad \{\text{since } y>0,\ y_2>y_1\}
\displaystyle =\int_{0}^{1}(4x-x^2)\,dx+\int_{1}^{\frac{5}{2}}\big[(4x-x^2)-(x^2-x)\big]\,dx
\displaystyle =\left[\frac{4x^2}{2}-\frac{x^3}{3}\right]_{0}^{1}+\int_{1}^{\frac{5}{2}}(5x-2x^2)\,dx
\displaystyle =\left[2x^2-\frac{x^3}{3}\right]_{0}^{1}+\left[\frac{5x^2}{2}-\frac{2x^3}{3}\right]_{1}^{\frac{5}{2}}
\displaystyle =\left(2-\frac{1}{3}\right)+\left[\frac{5}{2}\left(\frac{5}{2}\right)^2-\frac{2}{3}\left(\frac{5}{2}\right)^3-\frac{5}{2}+\frac{2}{3}\right]
\displaystyle =\frac{5}{3}+\left[\frac{125}{8}-\frac{125}{12}-\frac{15}{6}+\frac{4}{6}\right]
\displaystyle =\frac{5}{3}+\frac{55}{24}
\displaystyle =\frac{121}{24}\ \text{sq units}\quad\text{(1)}
\displaystyle \text{Area }(OCV'O)=\int_{0}^{1}|y|\,dx=\int_{0}^{1}(-y)\,dx \quad \{\text{since } y<0\}
\displaystyle =\int_{0}^{1}-(x^2-x)\,dx
\displaystyle =\int_{0}^{1}(x-x^2)\,dx
\displaystyle =\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_{0}^{1}
\displaystyle =\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\ \text{sq units}\quad\text{(2)}
\displaystyle \text{From (1) and (2)}
\displaystyle \text{Ratio}=\frac{\text{Area}(OBDCO)}{\text{Area}(OCV'O)}=\frac{\frac{121}{24}}{\frac{1}{6}}=\frac{121}{4}
\displaystyle \therefore \text{Required ratio }=121:4

\displaystyle \textbf{Question 50: }~\text{Find the area of the figure bounded by the curves } \\ y=\lvert x-1\rvert\text{ and }y=3-\lvert x\rvert.
\displaystyle \text{Answer:} \displaystyle \text{We have, }
\displaystyle y=|x-1|
\displaystyle \Rightarrow y=\begin{cases}x-1,&x\ge1\\1-x,&x<1\end{cases}
\displaystyle y=x-1\text{ is a straight line passing through }A(1,0)
\displaystyle y=1-x\text{ is a straight line passing through }A(1,0)\text{ and cutting the }y\text{-axis at }B(0,1)
\displaystyle y=3-|x|
\displaystyle \Rightarrow y=\begin{cases}3-x,&x\ge0\\3+x,&x<0\end{cases}
\displaystyle y=3-x\text{ is a straight line passing through }C(0,3)\text{ and }D(3,0)
\displaystyle y=3+x\text{ is a straight line passing through }C(0,3)\text{ and }D'(-3,0)
\displaystyle \text{The point of intersection is obtained by solving the simultaneous equations}
\displaystyle y=x-1
\displaystyle \text{and }y=3-x
\displaystyle \Rightarrow x-1=3-x
\displaystyle \Rightarrow 2x-4=0
\displaystyle \Rightarrow x=2
\displaystyle \Rightarrow y=2-1=1
\displaystyle \text{Thus }P(2,1)\text{ is the point of intersection of }y=x-1\text{ and }y=3-x
\displaystyle \text{Point of intersection for }y=1-x\text{ and }y=3+x
\displaystyle \Rightarrow 1-x=3+x
\displaystyle \Rightarrow 2x=-2
\displaystyle \Rightarrow x=-1
\displaystyle \Rightarrow y=1-(-1)=2
\displaystyle \text{Thus }Q(-1,2)\text{ is the point of intersection of }y=1-x\text{ and }y=3+x
\displaystyle \text{Since the character of the function changes at }C(0,3)\text{ and }A(1,0),\text{ draw }AM\perp x\text{-axis}
\displaystyle \text{Required area }=\text{ Shaded area }(QCPAQ)
\displaystyle =\text{Area }(QCB)+\text{Area }(BCMAB)+\text{Area }(AMPA)\quad\text{(1)}
\displaystyle \text{Area }(QCB)=\int_{-1}^{0}\big[(3+x)-(1-x)\big]\,dx
\displaystyle =\int_{-1}^{0}(2+2x)\,dx
\displaystyle =\left[2x+x^2\right]_{-1}^{0}
\displaystyle =0-(-2+1)=1\text{ sq. unit}\quad\text{(2)}
\displaystyle \text{Area }(BCMAB)=\int_{0}^{1}\big[(3-x)-(1-x)\big]\,dx
\displaystyle =\int_{0}^{1}2\,dx
\displaystyle =\left[2x\right]_{0}^{1}=2\text{ sq. units}\quad\text{(3)}
\displaystyle \text{Area }(AMPA)=\int_{1}^{2}\big[(3-x)-(x-1)\big]\,dx
\displaystyle =\int_{1}^{2}(4-2x)\,dx
\displaystyle =\left[4x-x^2\right]_{1}^{2}
\displaystyle =(8-4)-(4-1)=1\text{ sq. unit}\quad\text{(4)}
\displaystyle \text{From (1), (2), (3) and (4)}
\displaystyle \text{Shaded area }=1+2+1=4\text{ sq. units}

\displaystyle \textbf{Question 51: }~\text{If the area bounded by the parabola }y^{2}=4ax\text{ and the line } \\ y=mx\text{ is }\frac{a^{2}}{12}\text{ sq. units, then using integration, find the value of }m.
\displaystyle \text{Answer:} \displaystyle \text{The parabola }y^2=4ax\text{ opens towards the positive }x\text{-axis and its focus is }(a,0).
\displaystyle \text{The line }y=mx\text{ passes through the origin }(0,0).
\displaystyle \text{Solving }y^2=4ax\text{ and }y=mx,\text{ we get}
\displaystyle m^2x^2=4ax
\displaystyle \Rightarrow m^2x^2-4ax=0
\displaystyle \Rightarrow x(m^2x-4a)=0
\displaystyle \Rightarrow x=0\text{ or }x=\frac{4a}{m^2}
\displaystyle \text{So, the points of intersection of the given parabola and line are }O(0,0)\text{ and}
\displaystyle A\left(\frac{4a}{m^2},\frac{4a}{m}\right)
\displaystyle \text{Area bounded by the given parabola and line}
\displaystyle =\text{Area of the shaded region}
\displaystyle =\int_{0}^{\frac{4a}{m^2}} y_{\text{parabola}}\,dx-\int_{0}^{\frac{4a}{m^2}} y_{\text{line}}\,dx
\displaystyle =\int_{0}^{\frac{4a}{m^2}}\sqrt{4ax}\,dx-\int_{0}^{\frac{4a}{m^2}}mx\,dx
\displaystyle =\left[2\sqrt{a}\,\frac{x^{3/2}}{\frac{3}{2}}\right]_{0}^{\frac{4a}{m^2}}-\left[m\frac{x^2}{2}\right]_{0}^{\frac{4a}{m^2}}
\displaystyle =\frac{4\sqrt{a}}{3}\left(\frac{4a}{m^2}\right)^{3/2}-\frac{m}{2}\left(\frac{4a}{m^2}\right)^2
\displaystyle =\frac{32a^2}{3m^3}-\frac{8a^2}{m^3}
\displaystyle =\frac{8a^2}{3m^3}\text{ square units}
\displaystyle \text{But,}
\displaystyle \text{Area bounded by the given parabola and line }=\frac{a^2}{12}\text{ sq. units (Given)}
\displaystyle \Rightarrow \frac{8a^2}{3m^3}=\frac{a^2}{12}
\displaystyle \Rightarrow m^3=32
\displaystyle \Rightarrow m=\sqrt[3]{32}
\displaystyle \text{Thus, the value of }m\text{ is }\sqrt[3]{32}.

\displaystyle \textbf{Question 52: }~\text{If the area enclosed by the parabolas }y^{2}=16ax\text{ and } \\ x^{2}=16ay,\ a>0 \text{ is }\frac{1024}{3}\text{ square units, find the value of }a.
\displaystyle \text{Answer:} \displaystyle \text{The parabola } y^2 = 16ax \text{ opens towards the positive } x\text{-axis and its focus is } (4a,0).
\displaystyle \text{The parabola } x^2 = 16ay \text{ opens towards the positive } y\text{-axis and its focus is } (0,4a).
\displaystyle \text{Solving } y^2 = 16ax \text{ and } x^2 = 16ay, \text{ we get}
\displaystyle y=\frac{x^2}{16a}
\displaystyle \Rightarrow \left(\frac{x^2}{16a}\right)^2=16ax
\displaystyle \Rightarrow x^4=(16a)^3x
\displaystyle \Rightarrow x^4-(16a)^3x=0
\displaystyle \Rightarrow x\left[x^3-(16a)^3\right]=0
\displaystyle \Rightarrow x=0 \text{ or } x=16a
\displaystyle \text{So, the points of intersection of the given parabolas are } O(0,0) \text{ and } A(16a,16a).
\displaystyle \text{Area enclosed by the given parabolas}
\displaystyle = \text{Area of the shaded region}
\displaystyle = \int_{0}^{16a}\sqrt{16ax}\,dx-\int_{0}^{16a}\frac{x^2}{16a}\,dx
\displaystyle =4\sqrt{a}\int_{0}^{16a}x^{1/2}\,dx-\frac{1}{16a}\int_{0}^{16a}x^2\,dx
\displaystyle =4\sqrt{a}\left[\frac{2}{3}x^{3/2}\right]_{0}^{16a}-\frac{1}{16a}\left[\frac{x^3}{3}\right]_{0}^{16a}
\displaystyle =\frac{8\sqrt{a}}{3}(16a)^{3/2}-\frac{1}{48a}(16a)^3
\displaystyle =\frac{8\sqrt{a}}{3}\cdot64a\sqrt{a}-\frac{4096a^3}{48a}
\displaystyle =\frac{512a^2}{3}-\frac{256a^2}{3}
\displaystyle =\frac{256a^2}{3}\text{ square units}
\displaystyle \text{But, area enclosed by the given parabolas }=\frac{1024}{3}\text{ square units (Given)}
\displaystyle \Rightarrow \frac{256a^2}{3}=\frac{1024}{3}
\displaystyle \Rightarrow a^2=4
\displaystyle \Rightarrow a=2\quad(a>0)
\displaystyle \text{Thus, the value of } a \text{ is } 2.

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