Class 12: Areas of Bounded Regions – Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1: } \text{Using integration, find the area of the region bounded by the} \\ \text{following curves, after making a rough sketch: }y=1+|x+1|,\ x=-3, \\ x=3, y=0.\hspace{12.0cm} \text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have,}$
$\displaystyle y=1+|x+1| =\begin{cases} 1+(x+1), & x+1\ge0\\[2pt] 1-(x+1), & x+1<0 \end{cases} =\begin{cases} x+2, & x\ge-1\\ -x, & x<-1 \end{cases}$
$\displaystyle \text{Thus, the equations of the given curves are } \\ y=x+2\text{ for }x\ge-1,\ y=-x\text{ for }x<-1,\ x=-3,\ x=3\text{ and }y=0.$
$\displaystyle \text{Clearly, }y=x+2\text{ is a straight line cutting }x\text{- and }y\text{-axes at }(-2,0)\text{ and }(0,2)\text{ respectively.}$
$\displaystyle \text{The equation }y=-x\text{ for }x<-1\text{ represents that part of the line passing through the} \\ \text{origin and making an angle of }135^\circ\text{ with }x\text{-axis.}$
$\displaystyle \text{Clearly, }x=-3\text{ and }x=3\text{ are lines parallel to }y\text{-axis.}$

$\displaystyle \text{When we slice the shaded region into vertical strips, the strips change their character at }A.$
$\displaystyle \text{So, required area }=\text{Area }CDAB+\text{Area }ABEF.$
$\displaystyle \text{Area }CDAB:\ \text{each vertical strip has its upper end on }y=-x\text{ and lower end on }x\text{-axis.}$
$\displaystyle \Rightarrow \text{Area }CDAB=\int_{-3}^{-1}|y|\,dx=\int_{-3}^{-1}(-x)\,dx.$
$\displaystyle \text{Area }ABEF:\ \text{each vertical strip has its lower end on }x\text{-axis and upper end on }y=x+2.$
$\displaystyle \Rightarrow \text{Area }ABEF=\int_{-1}^{3}|y|\,dx=\int_{-1}^{3}(x+2)\,dx.$
$\displaystyle \text{Hence, required area }=\int_{-3}^{-1}(-x)\,dx+\int_{-1}^{3}(x+2)\,dx$
$\displaystyle =\left[-\frac{x^{2}}{2}\right]_{-3}^{-1}+\left[\frac{x^{2}}{2}+2x\right]_{-1}^{3}$
$\displaystyle =\left(-\frac{1}{2}+\frac{9}{2}\right)+\left(\frac{9}{2}+6-\frac{1}{2}+2\right)=16\ \text{sq.\ units.}$

$\displaystyle \textbf{Question 2: } \text{Using integration find the area of the triangle formed by positive} \\ x\text{-axis and tangent and normal to the circle }x^{2}+y^{2}=4\text{ at }(1,\sqrt{3}).\hspace{1.0cm} \text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the circle is}$
$\displaystyle x^{2}+y^{2}=4.$
$\displaystyle \text{Differentiating with respect to }x,\text{ we obtain}$
$\displaystyle 2x+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{x}{y}\Rightarrow \left(\frac{dy}{dx}\right)_{(1,\sqrt{3})}=-\frac{1}{\sqrt{3}}.$
$\displaystyle \text{So, the equations of tangent and normal to the circle at }(1,\sqrt{3})\text{ are}$
$\displaystyle y-\sqrt{3}=-\frac{1}{\sqrt{3}}(x-1)\ \ \text{and}\ \ y-\sqrt{3}=\sqrt{3}(x-1)\ \ \text{respectively.}$
$\displaystyle \Rightarrow x+\sqrt{3}\,y=4\ \ \text{and}\ \ y=\sqrt{3}\,x.$
$\displaystyle \text{The required triangular region is bounded by }y=0,\ y=\sqrt{3}x,\ \text{and }x+\sqrt{3}y=4.$

$\displaystyle \text{Let the required area be }A.$
$\displaystyle \text{For }0\le x\le 1,\ \text{upper boundary is }y_{1}=\sqrt{3}x.$
$\displaystyle \text{For }1\le x\le 4,\ \text{upper boundary is }x+\sqrt{3}y=4\Rightarrow y_{2}=\frac{4-x}{\sqrt{3}}.$
$\displaystyle A=\int_{0}^{1}y_{1}\,dx+\int_{1}^{4}y_{2}\,dx=\int_{0}^{1}\sqrt{3}x\,dx+\int_{1}^{4}\frac{4-x}{\sqrt{3}}\,dx.$
$\displaystyle A=\sqrt{3}\left[\frac{x^{2}}{2}\right]_{0}^{1}+\frac{1}{\sqrt{3}}\left[4x-\frac{x^{2}}{2}\right]_{1}^{4}=\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{3}}\left(8-\frac{7}{2}\right)=\frac{\sqrt{3}}{2}+\frac{9}{2\sqrt{3}}=2\sqrt{3}.$
$\displaystyle \text{Hence, the required area }=2\sqrt{3}\text{ sq. units.}$

$\displaystyle \textbf{Question 3: } \text{Find the area bounded by the curve }x^{2}=4y\text{ and the straight line } \\ x=4y-2. \hspace{4.0cm} \text{[CBSE 2004, 2005, 2010, 2013, 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equations of the given curves are}$
$\displaystyle x^{2}=4y\ \ \ ...(i)\qquad \text{and}\qquad x=4y-2\ \ \ ...(ii)$
$\displaystyle \text{From (ii), }y=\frac{x+2}{4}.$
$\displaystyle \text{To find the points of intersection, substitute in (i): }x^{2}=4\left(\frac{x+2}{4}\right)=x+2$
$\displaystyle \Rightarrow x^{2}-x-2=0\Rightarrow (x-2)(x+1)=0\Rightarrow x=2,-1.$
$\displaystyle \text{Corresponding }y\text{-values: }y=\frac{2+2}{4}=1,\ \ y=\frac{-1+2}{4}=\frac{1}{4}.$
$\displaystyle \text{Hence, the region lies between }x=-1\text{ and }x=2.$
$\displaystyle \text{For a vertical strip, upper curve is the line }y_{2}=\frac{x+2}{4} \\ \text{ and lower curve is the parabola }y_{1}=\frac{x^{2}}{4}.$

$\displaystyle A=\int_{-1}^{2}(y_{2}-y_{1})\,dx=\int_{-1}^{2}\left(\frac{x+2}{4}-\frac{x^{2}}{4}\right)\!dx$
$\displaystyle =\frac{1}{4}\int_{-1}^{2}(-x^{2}+x+2)\,dx=\frac{1}{4}\left[-\frac{x^{3}}{3}+\frac{x^{2}}{2}+2x\right]_{-1}^{2}$
$\displaystyle =\frac{1}{4}\left\{\left(-\frac{8}{3}+2+4\right)-\left(\frac{1}{3}+\frac{1}{2}-2\right)\right\}=\frac{1}{4}\cdot\frac{9}{2}=\frac{9}{8}.$
$\displaystyle \text{Hence, the required area }=\frac{9}{8}\text{ sq. units.}$

$\displaystyle \textbf{Question 4: } \text{Find the area of the region included between the parabolas } \\ y^{2}=4ax\text{ and }x^{2}=4ay,\text{ where }a>0. \hspace{3.0cm} \text{[CBSE 2003, 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equations of the given curves are}$
$\displaystyle y^{2}=4ax\ \ \ ...(i)\qquad \text{and}\qquad x^{2}=4ay\ \ \ ...(ii)$
$\displaystyle \text{To find the points of intersection, put }y=\frac{x^{2}}{4a}\text{ from (ii) into (i):}$
$\displaystyle \left(\frac{x^{2}}{4a}\right)^{2}=4ax\Rightarrow \frac{x^{4}}{16a^{2}}=4ax\Rightarrow x^{4}=64a^{3}x\Rightarrow x(x^{3}-64a^{3})=0$
$\displaystyle \Rightarrow x=0\ \text{or}\ x=4a$
$\displaystyle \text{From (ii): if }x=0\Rightarrow y=0,\ \text{and if }x=4a\Rightarrow y=\frac{(4a)^{2}}{4a}=4a.$

$\displaystyle \text{Hence, the curves intersect at }(0,0)\text{ and }(4a,4a).$
$\displaystyle \text{For a vertical strip }(dx),\text{ upper curve is from (i): }y_{2}=\sqrt{4ax}=2\sqrt{ax}, \\ \text{and lower curve is from (ii): }y_{1}=\frac{x^{2}}{4a}.$
$\displaystyle \text{Required area }A=\int_{0}^{4a}(y_{2}-y_{1})\,dx=\int_{0}^{4a}\left(2\sqrt{ax}-\frac{x^{2}}{4a}\right)\!dx$
$\displaystyle =\left[\frac{4\sqrt{a}}{3}x^{3/2}-\frac{x^{3}}{12a}\right]_{0}^{4a} =\frac{4\sqrt{a}}{3}(4a)^{3/2}-\frac{(4a)^{3}}{12a}$
$\displaystyle =\frac{4\sqrt{a}}{3}\cdot 8a^{3/2}-\frac{64a^{3}}{12a} =\frac{32a^{2}}{3}-\frac{16a^{2}}{3} =\frac{16a^{2}}{3}.$
$\displaystyle \text{Hence, the required area }=\frac{16a^{2}}{3}\text{ sq. units.}$

$\displaystyle \textbf{Question 5: } \text{Find the area of the region }\{(x,y):x^{2}\le y\le x\}. \hspace{0.0cm} \text{[CBSE 2005, 2011, 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{Let }R=\{(x,y):x^{2}\le y\le x\}.$
$\displaystyle \text{Here, }R=R_{1}\cap R_{2},\text{ where }R_{1}=\{(x,y):x^{2}\le y\}\text{ and }R_{2}=\{(x,y):y\le x\}.$
$\displaystyle \text{Clearly, }y=x^{2}\text{ is a parabola and }y=x\text{ is a straight line.}$

$\displaystyle \text{Solving }y=x^{2}\text{ and }y=x,\text{ we get }x^{2}=x\Rightarrow x(x-1)=0\Rightarrow x=0,1.$
$\displaystyle \text{Hence, the curves intersect at }(0,0)\text{ and }(1,1).$
$\displaystyle \text{For }0\le x\le1,\text{ upper curve is }y_{2}=x\text{ and lower curve is }y_{1}=x^{2}\ (y_{2}>y_{1}).$
$\displaystyle \text{Therefore, required area }A=\int_{0}^{1}(y_{2}-y_{1})\,dx=\int_{0}^{1}(x-x^{2})\,dx.$
$\displaystyle A=\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}.$
$\displaystyle \text{Hence, the required area }=\frac{1}{6}\text{ sq. unit.}$

$\displaystyle \textbf{Question 6: } \text{Find the area of the region }\{(x,y):x^{2}\le y\le |x|\}.\qquad \textbf{[CBSE 2002, 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }R=\{(x,y):x^{2}\le y\le |x|\}.$
$\displaystyle R=\{(x,y):x^{2}\le y\}\cap\{(x,y):y\le |x|\}.$
$\displaystyle R=\{(x,y):x^{2}\le y\}\cap\Big(\{(x,y):y\le x,\ x\ge0\}\cup\{(x,y):y\le -x,\ x<0\}\Big).$
$\displaystyle \text{Let }R_{1}=\{(x,y):x^{2}\le y\},\ R_{2}=\{(x,y):y\le x,\ x\ge0\},\ \\ R_{3}=\{(x,y):y\le -x,\ x<0\}.$ $\displaystyle \text{Curve }y=x^{2}\text{ is a parabola opening upward, and }y=\pm x\text{ are straight lines.}$
$\displaystyle \text{Since the region is symmetric about the }y\text{-axis, total area is twice the area in the first quadrant.}$
$\displaystyle \text{In the first quadrant, upper curve is }y_{2}=x\text{ and lower curve is }y_{1}=x^{2}.$
$\displaystyle A=2\int_{0}^{1}(y_{2}-y_{1})\,dx =2\int_{0}^{1}(x-x^{2})\,dx.$
$\displaystyle A=2\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1} =2\left(\frac12-\frac13\right) =2\cdot\frac16=\frac13.$
$\displaystyle \text{Hence, the required area }=\frac{1}{3}\text{ sq. units.}$

$\displaystyle \textbf{Question 7: } \text{Find the area of the region:} \\ \{(x,y):0\le y\le x^{2}+1,\ 0\le y\le x+1,\ 0\le x\le2\}. \hspace{1.0cm} \text{[CBSE 2001C]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }R=\{(x,y):0\le y\le x^{2}+1,\ 0\le y\le x+1,\ 0\le x\le2\}.$
$\displaystyle R=\{(x,y):0\le y\le x^{2}+1\}\cap\{(x,y):0\le y\le x+1\}\cap\{(x,y):0\le x\le2\}.$
$\displaystyle \Rightarrow R=R_{1}\cap R_{2}\cap R_{3},$
$\displaystyle \text{where }R_{1}=\{(x,y):0\le y\le x^{2}+1\},\ R_{2}=\{(x,y):0\le y\le x+1\},\ R_{3}=\{(x,y):0\le x\le2\}.$
$\displaystyle \text{Region }R_{1}:\ y=x^{2}+1\text{ is a parabola with vertex at }(0,1)\text{ opening upwards.}$
$\displaystyle \text{Region }R_{2}:\ y=x+1\text{ is a straight line.}$
$\displaystyle \text{Region }R_{3}:\ 0\le x\le2.$
$\displaystyle \text{Solving }y=x^{2}+1\text{ and }y=x+1,\text{ we get intersections at }A(0,1)\text{ and }B(1,2).$
$\displaystyle \text{Divide the region at }x=1.$
$\displaystyle \text{Area }OABCO=\int_{0}^{1}(x^{2}+1)\,dx.$
$\displaystyle \text{Area }BDECB=\int_{1}^{2}(x+1)\,dx.$
$\displaystyle \text{Required area }=\int_{0}^{1}(x^{2}+1)\,dx+\int_{1}^{2}(x+1)\,dx.$
$\displaystyle =\left[\frac{x^{3}}{3}+x\right]_{0}^{1}+\left[\frac{x^{2}}{2}+x\right]_{1}^{2}.$
$\displaystyle =\frac{4}{3}+\frac{5}{2}=\frac{23}{6}.$
$\displaystyle \text{Hence, the required area }=\frac{23}{6}\text{ sq. units.}$

$\displaystyle \textbf{Question 8: } \text{Find the area of the region }\{(x,y):x^{2}+y^{2}\le1\le x+y\}. \hspace{2.0cm} \text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }R=\{(x,y):x^{2}+y^{2}\le1\le x+y\}.$
$\displaystyle \text{Then, }R=\{(x,y):x^{2}+y^{2}\le1\}\cap\{(x,y):1\le x+y\}=R_{1}\cap R_{2},$
$\displaystyle \text{where }R_{1}=\{(x,y):x^{2}+y^{2}\le1\}\text{ and }R_{2}=\{(x,y):1\le x+y\}.$
$\displaystyle \text{Region }R_{1}:\ x^{2}+y^{2}=1\text{ is a circle with centre }(0,0)\text{ and radius }1.$
$\displaystyle \text{Hence }x^{2}+y^{2}\le1\text{ represents the interior of the circle.}$
$\displaystyle \text{Region }R_{2}:\ x+y=1\text{ is a straight line cutting the axes at }(1,0)\text{ and }(0,1).$
$\displaystyle \text{Since }(0,0)\text{ does not satisfy }x+y\ge1,\text{ the required region lies above the line.}$
$\displaystyle \text{Let us slice the region into vertical strips. For a strip at }x,$
$\displaystyle \text{lower boundary }y_{1}=1-x,\text{ upper boundary }y_{2}=\sqrt{1-x^{2}}.$
$\displaystyle \text{Therefore, area }A=\int_{0}^{1}(y_{2}-y_{1})\,dx=\int_{0}^{1}\!\left[\sqrt{1-x^{2}}-(1-x)\right]dx.$
$\displaystyle \text{Now, }A=\left[\frac{1}{2}x\sqrt{1-x^{2}}+\frac{1}{2}\sin^{-1}x-x+\frac{x^{2}}{2}\right]_{0}^{1}.$
$\displaystyle \text{Hence, }A=\left(0+\frac{\pi}{4}-1+\frac{1}{2}\right)-0=\left(\frac{\pi}{4}-\frac{1}{2}\right)\text{ sq. units.}$

$\displaystyle \text{Question 9: } \text{Find the area of the region }\{(x,y):y^{2}\le4x,\ 4x^{2}+4y^{2}\le9\}. \hspace{0.0cm} \text{[CBSE 2017]} \\ {\hspace{7.0cm} \text{or}}$
$\displaystyle \text{Find the area of the circle }4x^{2}+4y^{2}=9\text{ which is interior to the parabola } \\ y^{2}=4x. \hspace{11.0cm} \text{[CBSE 2010, 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }R=\{(x,y):y^{2}\le4x,\ 4x^{2}+4y^{2}\le9\}.$
$\displaystyle R=\{(x,y):y^{2}\le4x\}\cap\{(x,y):4x^{2}+4y^{2}\le9\}.$
$\displaystyle \Rightarrow R=R_{1}\cap R_{2},\text{ where }R_{1}=\{(x,y):y^{2}\le4x\} \\ \text{ and }R_{2}=\{(x,y):4x^{2}+4y^{2}\le9\}.$
$\displaystyle \text{Region }R_{1}:\ y^{2}=4x\text{ is a parabola with vertex at the origin and axis along }x\text{-axis.} \\ \text{Thus }y^{2}\le4x\text{ is the region inside the parabola.}$
$\displaystyle \text{Region }R_{2}:\ 4x^{2}+4y^{2}=9\Rightarrow x^{2}+y^{2}=\left(\frac{3}{2}\right)^{2} \text{ is a circle with centre at origin and radius }\frac{3}{2}.$
$\displaystyle \text{Hence the region }R\text{ is bounded by the parabola }y^{2}=4x\text{ and the circle } \\ x^{2}+y^{2}=\left(\frac{3}{2}\right)^{2}.$
$\displaystyle \text{To find points of intersection, solve }y^{2}=4x\text{ and }4x^{2}+4y^{2}=9.$
$\displaystyle 4x^{2}+16x-9=0\Rightarrow (2x+9)(2x-1)=0\Rightarrow x=\frac12,-\frac92.$
$\displaystyle \text{Only }x=\frac12\text{ is admissible, giving }y=\pm\sqrt{2}. \text{ Hence curves intersect at }\left(\frac12,\pm\sqrt{2}\right).$
$\displaystyle \text{By symmetry about }x\text{-axis, required area }A=2(\text{area above }x\text{-axis}).$
$\displaystyle A=2\left[\int_{0}^{1/2}2\sqrt{x}\,dx+\int_{1/2}^{3/2}\sqrt{\frac94-x^{2}}\,dx\right].$
$\displaystyle A=\frac{2\sqrt{2}}{3}+\frac{9\pi}{8}-\frac94\sin^{-1}\!\left(\frac13\right)\text{ sq. units.}$

$\displaystyle \textbf{Question 10: } \text{Find the area of the region }\{(x,y):x^{2}+y^{2}\le2ax,\ y^{2}\ge ax,\ x\ge0,\ y\ge0\}.\qquad \text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }R=\{(x,y):x^{2}+y^{2}\le2ax,\ y^{2}\ge ax,\ x\ge0,\ y\ge0\}.$
$\displaystyle R=\{(x,y):x^{2}+y^{2}\le2ax\}\cap\{(x,y):y^{2}\ge ax\}\cap\{(x,y):x\ge0,y\ge0\}.$
$\displaystyle \Rightarrow R=R_{1}\cap R_{2}\cap R_{3},\text{ where }R_{1}=\{(x,y):x^{2}+y^{2}\le2ax\}, \\ R_{2}=\{(x,y):y^{2}\ge ax\}\text{ and }R_{3}=\{(x,y):x\ge0,y\ge0\}.$
$\displaystyle \text{Region }R_{1}:\ x^{2}+y^{2}=2ax\Rightarrow (x-a)^{2}+y^{2}=a^{2},\text{ a circle with centre }(a,0)\text{ and radius }a.$
$\displaystyle \text{Region }R_{2}:\ y^{2}=ax\text{ is a parabola with vertex at }(0,0).$
$\displaystyle \text{Region }R_{3}:\ x\ge0,\ y\ge0\text{ gives the first quadrant.}$
$\displaystyle \text{The curves are }x^{2}+y^{2}=2ax\ \text{and}\ y^{2}=ax.$
$\displaystyle \text{Solving simultaneously gives intersection points }(0,0)\text{ and }(a,a).$
$\displaystyle \text{Using vertical strips, lower curve is }y_{1}=\sqrt{ax},\text{ upper curve satisfies } \\ (x-a)^{2}+y_{2}^{2}=a^{2}.$
$\displaystyle A=\int_{0}^{a}(y_{2}-y_{1})\,dx=\int_{0}^{a}\!\left[\sqrt{a^{2}-(x-a)^{2}}-\sqrt{ax}\right]dx.$
$\displaystyle A=\left[\frac12(x-a)\sqrt{a^{2}-(x-a)^{2}}+\frac12a^{2}\sin^{-1}\!\left(\frac{x-a}{a}\right)-\frac23a^{1/2}x^{3/2}\right]_{0}^{a}.$
$\displaystyle A=-\frac23a^{2}-\frac12a^{2}\!\left(-\frac{\pi}{2}\right)=\left(\frac{\pi}{4}-\frac23\right)a^{2}\text{ sq. units.}$

$\displaystyle \textbf{Question 11: } \text{Find the area of the region enclosed between the two circles } \\ x^{2}+y^{2}=1\text{ and }(x-1)^{2}+y^{2}=1.\hspace{6.0cm} \text{[CBSE 2007, 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equations of the given circles are}$
$\displaystyle x^{2}+y^{2}=1\qquad ...(i)$
$\displaystyle (x-1)^{2}+(y-0)^{2}=1\qquad ...(ii)$
$\displaystyle \text{Clearly, (i) represents a circle with centre }(0,0)\text{ and radius }1.$
$\displaystyle \text{Equation (ii) represents a circle with centre }(1,0)\text{ and radius }1.$

$\displaystyle \text{To find the points of intersection, solve (i) and (ii) simultaneously.}$
$\displaystyle \text{The two curves intersect at }A\!\left(\frac12,\frac{\sqrt3}{2}\right)\text{ and }D\!\left(\frac12,-\frac{\sqrt3}{2}\right).$
$\displaystyle \text{Since the curves are symmetric about the }x\text{-axis, required area }=2(\text{Area }OABCO).$
$\displaystyle \text{Divide }OABCO\text{ into two parts: }OACO\text{ and }CABC.$
$\displaystyle \text{Area }OACO=\int_{0}^{1/2}y_{1}\,dx,\qquad y_{1}=\sqrt{1-(x-1)^{2}}.$
$\displaystyle \text{Area }CABC=\int_{1/2}^{1}y_{2}\,dx,\qquad y_{2}=\sqrt{1-x^{2}}.$
$\displaystyle A=2\!\left[\int_{0}^{1/2}\sqrt{1-(x-1)^{2}}\,dx+\int_{1/2}^{1}\sqrt{1-x^{2}}\,dx\right].$
$\displaystyle A=2\!\left[\frac12(x-1)\sqrt{1-(x-1)^{2}}+\frac12\sin^{-1}(x-1)\right]_{0}^{1/2} +2\!\left[\frac12x\sqrt{1-x^{2}}+\frac12\sin^{-1}x\right]_{1/2}^{1}.$
$\displaystyle A=2\!\left[-\frac{\sqrt3}{4}+\sin^{-1}\!\left(-\frac12\right)-\sin^{-1}(-1)\right] +2\!\left[\sin^{-1}(1)-\frac{\sqrt3}{4}-\sin^{-1}\!\left(\frac12\right)\right].$
$\displaystyle A=\frac{2\pi}{3}-\frac{\sqrt3}{2}\ \text{sq. units.}$

$\displaystyle \text{Question 12: } \text{Using integration, find the area of triangle }ABC\text{ whose vertices are } \\ A(2,5),\ B(4,7)\text{ and }C(6,2).\hspace{5.0cm} \text{[CBSE 2001, 2010, 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{First, find the equations of the sides.}$
$\displaystyle \text{Side }AB:\ \frac{y-5}{7-5}=\frac{x-2}{4-2}\Rightarrow y-5=x-2\Rightarrow x-y+3=0\ \ \ ...(i)$
$\displaystyle \text{Side }BC:\ \frac{y-7}{2-7}=\frac{x-4}{6-4}\Rightarrow 2(y-7)=-5(x-4)\Rightarrow 5x+2y-34=0\ \ \ ...(ii)$
$\displaystyle \text{Side }AC:\ \frac{y-5}{2-5}=\frac{x-2}{6-2}\Rightarrow 4(y-5)=-3(x-2)\Rightarrow 3x+4y-26=0\ \ \ ...(iii)$
$\displaystyle \text{Clearly, Area}(\triangle ABC)=\text{Area}(ADB)+\text{Area}(BDC),\text{ where we split the region at }x=4.$
$\displaystyle \text{From (i), }y_2=x+3\text{ on }AB,\ \text{and from (iii), }y_1=\frac{26-3x}{4}\text{ on }AC.$
$\displaystyle \text{Area }ADB=\int_{2}^{4}(y_2-y_1)\,dx \\ =\int_{2}^{4}\left[(x+3)-\frac{26-3x}{4}\right]dx \\ =\frac{1}{4}\int_{2}^{4}(7x-14)\,dx.$
$\displaystyle \text{From (ii), }y_4=\frac{34-5x}{2}\text{ on }BC,\ \text{and from (iii), }y_3=\frac{26-3x}{4}\text{ on }AC.$
$\displaystyle \text{Area }BDC=\int_{4}^{6}(y_4-y_3)\,dx \\ =\int_{4}^{6}\left[\frac{34-5x}{2}-\frac{26-3x}{4}\right]dx \\ =\frac{1}{4}\int_{4}^{6}(42-7x)\,dx.$
$\displaystyle \text{Now, }\int_{2}^{4}(7x-14)\,dx=\left[\frac{7}{2}x^{2}-14x\right]_{2}^{4}=14, \\ \int_{4}^{6}(42-7x)\,dx=\left[42x-\frac{7}{2}x^{2}\right]_{4}^{6}=14.$
$\displaystyle \text{Hence, Area }ADB=\frac{1}{4}\cdot 14=\frac{7}{2},\ \ \text{Area }BDC=\frac{1}{4}\cdot 14=\frac{7}{2}.$
$\displaystyle \text{Therefore, Area}(\triangle ABC)=\frac{7}{2}+\frac{7}{2}=7\text{ sq. units.}$

$\displaystyle \textbf{Question 13: } \text{Prove that the curves }y^{2}=4x\text{ and }x^{2}=4y \text{ divide the area of the square} \\ \text{bounded by }x=0,\ y=0,\ x=4,\ y=4\text{ into three equal parts.}\hspace{0.0cm} \text{[CBSE 2009, 2015, 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A_{1},A_{2},A_{3}\text{ denote the areas of the regions }OAPLO,\ OAPBO\text{ and }OBPMO \\ \text{respectively}.$

$\displaystyle \text{We have to prove that }A_{1}=A_{2}=A_{3}.$
$\displaystyle \text{Area }A_{1}\text{ (region under }x^{2}=4y\text{ from }x=0\text{ to }x=4):$
$\displaystyle x^{2}=4y\Rightarrow y=\frac{x^{2}}{4}.$
$\displaystyle A_{1}=\int_{0}^{4}y\,dx=\int_{0}^{4}\frac{x^{2}}{4}\,dx=\frac{1}{4}\left[\frac{x^{3}}{3}\right]_{0}^{4}=\frac{1}{4}\cdot\frac{64}{3}=\frac{16}{3}.$
$\displaystyle \text{Area }A_{2}\text{ (region between }y^{2}=4x\text{ and }x^{2}=4y\text{ from }x=0\text{ to }x=4):$
$\displaystyle y^{2}=4x\Rightarrow y=2\sqrt{x},\qquad x^{2}=4y\Rightarrow y=\frac{x^{2}}{4}.$
$\displaystyle A_{2}=\int_{0}^{4}\left(2\sqrt{x}-\frac{x^{2}}{4}\right)\!dx=\left[\frac{4}{3}x^{3/2}-\frac{x^{3}}{12}\right]_{0}^{4}=\left(\frac{4}{3}\cdot 8-\frac{64}{12}\right)=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}.$
$\displaystyle \text{Area }A_{3}\text{ (region to the right of }y^{2}=4x\text{ between }y=0\text{ and }y=4):$
$\displaystyle y^{2}=4x\Rightarrow x=\frac{y^{2}}{4}.$
$\displaystyle A_{3}=\int_{0}^{4}x\,dy=\int_{0}^{4}\frac{y^{2}}{4}\,dy=\frac{1}{4}\left[\frac{y^{3}}{3}\right]_{0}^{4}=\frac{1}{4}\cdot\frac{64}{3}=\frac{16}{3}.$
$\displaystyle \text{Hence, }A_{1}=A_{2}=A_{3}=\frac{16}{3}\text{ sq. units, so the two curves divide the square into} \\ \text{three equal parts.}$

$\displaystyle \text{Question 14: } \text{Sketch the region common to the circle }x^{2}+y^{2}=16\text{ and the parabola } \\ x^{2}=6y.\text{ Also, find its area using integration.}\hspace{4.0cm} \text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equations of the given curves are}$
$\displaystyle x^{2}+y^{2}=16\ \ \ ...(i)\qquad \text{and}\qquad x^{2}=6y\ \ \ ...(ii)$
$\displaystyle \text{Clearly, (i) is a circle with centre }(0,0)\text{ and radius }4,\text{ and (ii) is a parabola opening} \\ \text{upward with vertex at }(0,0).$
$\displaystyle \text{To find the points of intersection, put }y=\frac{x^{2}}{6}\text{ from (ii) into (i):}$
$\displaystyle x^{2}+\left(\frac{x^{2}}{6}\right)^{2}=16\Rightarrow x^{2}+\frac{x^{4}}{36}=16\Rightarrow x^{4}+36x^{2}-576=0$
$\displaystyle \Rightarrow (x^{2}+48)(x^{2}-12)=0\Rightarrow x^{2}=12\Rightarrow x=\pm 2\sqrt{3}.$
$\displaystyle \text{Putting }x=\pm 2\sqrt{3}\text{ in (ii), we get }y=\frac{12}{6}=2.$
$\displaystyle \text{Thus, the curves intersect at }A(2\sqrt{3},2)\text{ and }C(-2\sqrt{3},2). \text{ The common region is symmetric} \\ \text{about the }y\text{-axis.}$
$\displaystyle \text{Required area }=2\times(\text{area of the part in the first quadrant}).$
$\displaystyle \text{We slice the region into horizontal strips. For }0\le y\le 2,\text{ the right boundary is the} \\ \text{parabola }x_{1}=\sqrt{6y}.$
$\displaystyle \text{For }2\le y\le 4,\text{ the right boundary is the circle }x_{2}=\sqrt{16-y^{2}}.$
$\displaystyle \text{Area in first quadrant }=\int_{0}^{2}x_{1}\,dy+\int_{2}^{4}x_{2}\,dy=\int_{0}^{2}\sqrt{6y}\,dy+\int_{2}^{4}\sqrt{16-y^{2}}\,dy.$
$\displaystyle \text{Hence, required area }=2\left[\int_{0}^{2}\sqrt{6y}\,dy+\int_{2}^{4}\sqrt{16-y^{2}}\,dy\right].$
$\displaystyle \int_{0}^{2}\sqrt{6y}\,dy=\sqrt{6}\int_{0}^{2}y^{1/2}\,dy=\sqrt{6}\left[\frac{2}{3}y^{3/2}\right]_{0}^{2}=\frac{8\sqrt{3}}{3}.$
$\displaystyle \int \sqrt{16-y^{2}}\,dy=\frac{1}{2}\left[y\sqrt{16-y^{2}}+16\sin^{-1}\left(\frac{y}{4}\right)\right]+C.$
$\displaystyle \int_{2}^{4}\sqrt{16-y^{2}}\,dy=\left[\frac{1}{2}\left(y\sqrt{16-y^{2}}+16\sin^{-1}\left(\frac{y}{4}\right)\right)\right]_{2}^{4}$
$\displaystyle =\left[8\sin^{-1}(1)\right]-\left[\sqrt{12}+8\sin^{-1}\left(\frac{1}{2}\right)\right]=4\pi-2\sqrt{3}-\frac{4\pi}{3}= \frac{8\pi}{3}-2\sqrt{3}.$
$\displaystyle \text{So, required area }=2\left[\frac{8\sqrt{3}}{3}+\left(\frac{8\pi}{3}-2\sqrt{3}\right)\right]=2\left[\frac{2\sqrt{3}}{3}+\frac{8\pi}{3}\right] \\ =\frac{4\sqrt{3}}{3}+\frac{16\pi}{3}\ \text{sq. units.}$