Class 12: Application of Integrals – ISC Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1: } \text{Using integration, find the area bounded by the curve } y^{2}=4ax \\ \text{ and the line } x=a. \hspace{2.0cm} \text{ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Shaded area is enclosed by }y^{2}=4ax\text{ and }x=a$
$\displaystyle \text{Required area}=2\int_{0}^{a}y\,dx=2\int_{0}^{a}\sqrt{4ax}\,dx$
$\displaystyle =4\sqrt{a}\int_{0}^{a}\sqrt{x}\,dx$
$\displaystyle =4\sqrt{a}\left[\frac{x^{3/2}}{3/2}\right]_{0}^{a}$
$\displaystyle =4\sqrt{a}\times\frac{2}{3}\left[a^{3/2}-0\right]$
$\displaystyle =\frac{8\sqrt{a}}{3}\cdot a^{3/2}$
$\displaystyle =\frac{8}{3}a^{2}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 2: } \text{Using integration, find the area of the region bounded by the curves }\\ y^{2}=4x \text{ and } x^{2}=4y. \hspace{2.0cm} \text{ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have two parabolas, }y^{2}=4x\text{ and }x^{2}=4y$
$\displaystyle \text{Since }y^{2}=4x$
$\displaystyle \Rightarrow \left(\frac{x^{2}}{4}\right)^{2}=4x\quad\left[\because \frac{x^{2}}{4}=y\right]$
$\displaystyle \Rightarrow \frac{x^{4}}{16}=4x$
$\displaystyle \Rightarrow x^{4}-64x=0$
$\displaystyle \Rightarrow x(x^{3}-64)=0$
$\displaystyle \Rightarrow x=0\text{ or }x^{3}=64$
$\displaystyle \Rightarrow x=0\text{ or }x=4$
$\displaystyle \text{When }x=0,\ y=0$
$\displaystyle \text{When }x=4,\ y=4$
$\displaystyle \therefore \text{Required area}=\int_{0}^{4}\left[(y\text{ of }y^{2}=4x)-(y\text{ of }x^{2}=4y)\right]dx$
$\displaystyle =\int_{0}^{4}\left(2\sqrt{x}-\frac{x^{2}}{4}\right)dx$
$\displaystyle =\left[2\cdot\frac{2}{3}x^{3/2}-\frac{1}{4}\cdot\frac{x^{3}}{3}\right]_{0}^{4}$
$\displaystyle =\left[\frac{4}{3}x^{3/2}-\frac{x^{3}}{12}\right]_{0}^{4}$
$\displaystyle =\frac{4}{3}(4^{3/2}-0)-\frac{1}{12}(4^{3}-0)$
$\displaystyle =\frac{4}{3}(8)-\frac{64}{12}$
$\displaystyle =\frac{32}{3}-\frac{16}{3}=\frac{16}{3}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 3: } \text{Find the area of the region bounded by the curve } x^{2}=4y \\ \text{ and the line } x=4y-2. \hspace{2.0cm} \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The area bounded by }x^{2}=4y\text{ and }x=4y-2\text{ is represented by shaded area}$
$\displaystyle \text{Let }A\text{ and }B\text{ be intersection points of line and parabola}$
$\displaystyle A\left(-1,\frac{1}{4}\right)\text{ and }B(2,1)$
$\displaystyle \therefore \text{Area of shaded region}$
$\displaystyle =\text{ar}(ABCDA)-\text{ar}(OBCDAO)$
$\displaystyle =\int_{-1}^{2}(y_{\text{line}}-y_{\text{parabola}})\,dx$
$\displaystyle =\int_{-1}^{2}\left(\frac{x+2}{4}-\frac{x^{2}}{4}\right)dx$
$\displaystyle =\frac{1}{4}\left[\frac{x^{2}}{2}+2x-\frac{x^{3}}{3}\right]_{-1}^{2}$
$\displaystyle =\frac{1}{4}\left[\left(2+4-\frac{8}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)\right]$
$\displaystyle =\frac{1}{4}\left[\frac{10}{3}-\left(-\frac{7}{6}\right)\right]$
$\displaystyle =\frac{1}{4}\left(\frac{20+7}{6}\right)=\frac{27}{24}=\frac{9}{8}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 4: } \text{Sketch the graph of the curves } y^{2}=x \text{ and } y^{2}=4-3x \\ \text{ and find the area enclosed between them.} \hspace{2.0cm} \text{ ISC 2020, 2014, 2007}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given curves are }y^{2}=x \quad ...(i)$
$\displaystyle \text{and }y^{2}=4-3x \quad ...(ii)$
$\displaystyle \text{For }y^{2}=4-3x,\text{ vertex is }A\left(\frac{4}{3},0\right)$
$\displaystyle \text{and it intersects }Y\text{-axis at }P(0,2)\text{ and }Q(0,-2)$ $\displaystyle \text{Also, }y^{2}=x\text{ has vertex }O(0,0)\text{ and opens towards right}$
$\displaystyle \text{From Eqs. (i) and (ii), }x=4-3x$
$\displaystyle \Rightarrow 4x=4\Rightarrow x=1$
$\displaystyle \text{From Eq. (i), }y^{2}=1\Rightarrow y=\pm1$
$\displaystyle \text{Points of intersection are }R(1,1)\text{ and }S(1,-1)$
$\displaystyle \text{Area of curve }ORTO=\int_{0}^{1}y\,dx=\int_{0}^{1}\sqrt{x}\,dx$
$\displaystyle =\left[\frac{x^{3/2}}{3/2}\right]_{0}^{1}=\frac{2}{3}$
$\displaystyle \text{Area of curve }RATR=\int_{1}^{4/3}y\,dx=\int_{1}^{4/3}\sqrt{4-3x}\,dx$
$\displaystyle =\left[-\frac{2}{9}(4-3x)^{3/2}\right]_{1}^{4/3}$
$\displaystyle =-\frac{2}{9}\left[(0)^{3/2}-(1)^{3/2}\right]=\frac{2}{9}$
$\displaystyle \text{Area of curve }ORATO=\frac{2}{3}+\frac{2}{9}=\frac{8}{9}$
$\displaystyle \therefore \text{Required area }ORASO=2\times\text{Area of curve }ORATO$
$\displaystyle =2\times\frac{8}{9}=\frac{16}{9}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 5: } \text{Draw a rough sketch and find the area bounded by the curve } \\ x^{2}=y \text{ and } x+y=2. \hspace{2.0cm} \text{ ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, equation of curve }x^{2}=y\text{ is a parabola}$
$\displaystyle \text{with vertex }(0,0)\text{ and it opens upward}$
$\displaystyle \text{Also, equation of line is }x+y=2$
$\displaystyle \text{The line intercepts coordinate axes at }(2,0)\text{ and }(0,2)$
$\displaystyle \text{Intersection of curve and line is}$
$\displaystyle x+x^{2}=2$
$\displaystyle \Rightarrow x^{2}+x-2=0$
$\displaystyle \Rightarrow x^{2}+2x-x-2=0$
$\displaystyle \Rightarrow x(x+2)-1(x+2)=0$
$\displaystyle \Rightarrow (x-1)(x+2)=0$
$\displaystyle \Rightarrow x=1,-2$
$\displaystyle \text{When }x=1,\ y=1^{2}=1$
$\displaystyle \text{When }x=-2,\ y=(-2)^{2}=4$
$\displaystyle \text{Hence, points of intersection are }A(1,1)\text{ and }B(-2,4)$
$\displaystyle \therefore \text{Required area}=\text{Area of shaded region }(AOBDA)$
$\displaystyle =\int_{-2}^{1}\left[y_{\text{line}}-y_{\text{parabola}}\right]dx$
$\displaystyle =\int_{-2}^{1}\left[(2-x)-x^{2}\right]dx$
$\displaystyle =\int_{-2}^{1}(2-x-x^{2})dx$
$\displaystyle =\left[2x-\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{-2}^{1}$
$\displaystyle =2-\frac{1}{2}-\frac{1}{3}-\left(-4-\frac{4}{2}+\frac{8}{3}\right)$
$\displaystyle =\frac{7}{6}-\left(-6+\frac{8}{3}\right)$
$\displaystyle =\frac{7}{6}+6-\frac{8}{3}$
$\displaystyle =\frac{7+36-16}{6}=\frac{27}{6}=\frac{9}{2}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 6: } \text{Draw a rough sketch of the curve and find the area of the region} \\ \text{bounded by the curve } y^{2}=8x \text{ and the line } x=2. \hspace{2.0cm} \text{ ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, curve }y^{2}=8x\text{ and line }x=2$
$\displaystyle \text{Let }B\text{ and }C\text{ be intersection points of line and curve}$
$\displaystyle B(2,4)\text{ and }C(2,-4)$
$\displaystyle \text{Area bounded by }y^{2}=8x\text{ and }x=2$
$\displaystyle =2(\text{area of region }OABO)$
$\displaystyle =2\int_{0}^{2}y\,dx=2\int_{0}^{2}\sqrt{8x}\,dx$
$\displaystyle =2\cdot2\sqrt{2}\int_{0}^{2}\sqrt{x}\,dx$
$\displaystyle =4\sqrt{2}\left[\frac{2}{3}x^{3/2}\right]_{0}^{2}$
$\displaystyle =\frac{8\sqrt{2}}{3}(2^{3/2}-0)$
$\displaystyle =\frac{8\sqrt{2}}{3}\times2\sqrt{2}=\frac{32}{3}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 7: } \text{Draw a rough sketch of the curve } y^{2}=4x \text{ and find the area of} \\ \text{the region enclosed by the curve and the line } y=x. \hspace{2.0cm} \text{ISC 2017}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equations are }y^{2}=4x \quad ...(i)$
$\displaystyle \text{and }y=x \quad ...(ii)$
$\displaystyle \text{On putting }y=x\text{ in Eq. (i), we get}$
$\displaystyle x^{2}=4x$
$\displaystyle \Rightarrow x(x-4)=0$
$\displaystyle \Rightarrow x=0\text{ or }x=4$
$\displaystyle \text{Thus, points of intersection are } \\ O(0,0)\text{ and }A(4,4)$
$\displaystyle \text{Required area}=\int_{0}^{4}\left[y_{\text{parabola}}-y_{\text{line}}\right]dx$
$\displaystyle =\int_{0}^{4}\left(2\sqrt{x}-x\right)dx$
$\displaystyle =\left[2\cdot\frac{2}{3}x^{3/2}-\frac{x^{2}}{2}\right]_{0}^{4}$
$\displaystyle =\left[\frac{4}{3}x^{3/2}-\frac{x^{2}}{2}\right]_{0}^{4}$
$\displaystyle =\frac{4}{3}(4^{3/2})-\frac{16}{2}$
$\displaystyle =\frac{4}{3}\cdot8-8=\frac{32}{3}-8=\frac{8}{3}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 8: } \text{Find the area of the region bounded by the curves } \\ y=6x-x^{2} \text{ and } y=x^{2}-2x. \hspace{2.0cm} \text{ ISC 2016}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given curves are }y=6x-x^{2} \quad ...(i)$
$\displaystyle \text{and }y=x^{2}-2x \quad ...(ii)$
$\displaystyle \text{On solving, }6x-x^{2}=x^{2}-2x$
$\displaystyle \Rightarrow 2x^{2}-8x=0$
$\displaystyle \Rightarrow x(x-4)=0$
$\displaystyle \Rightarrow x=0,4$
$\displaystyle \text{Thus, points of intersection are } \\ (0,0)\text{ and }(4,8)$
$\displaystyle \text{Required area}=\int_{0}^{4}\left[(6x-x^{2})-(x^{2}-2x)\right]dx$
$\displaystyle =\int_{0}^{4}(8x-2x^{2})dx$
$\displaystyle =\left[8\cdot\frac{x^{2}}{2}-2\cdot\frac{x^{3}}{3}\right]_{0}^{4}$
$\displaystyle =\left[4x^{2}-\frac{2x^{3}}{3}\right]_{0}^{4}$
$\displaystyle =64-\frac{128}{3}=\frac{64}{3}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 9: } \text{Find the smaller area enclosed by the circle } x^{2}+y^{2}=4 \\ \text{ and the line } x+y=2. \hspace{2.0cm} \text{ ISC 2015}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, equation of circle is }x^{2}+y^{2}=4$
$\displaystyle \text{with centre }(0,0)\text{ and radius }2$
$\displaystyle \text{and equation of line is }x+y=2$
$\displaystyle \text{Putting }y=2-x\text{ in circle, we get}$
$\displaystyle x^{2}+(2-x)^{2}=4$
$\displaystyle \Rightarrow x^{2}+4+x^{2}-4x=4$
$\displaystyle \Rightarrow 2x^{2}-4x=0$
$\displaystyle \Rightarrow 2x(x-2)=0$
$\displaystyle \Rightarrow x=0,2$
$\displaystyle \text{Points of intersection are }(0,2)\text{ and }(2,0)$
$\displaystyle \text{Required area}=\int_{0}^{2}\left[\sqrt{4-x^{2}}-(2-x)\right]dx$
$\displaystyle =\int_{0}^{2}\sqrt{4-x^{2}}\,dx-\int_{0}^{2}(2-x)\,dx$
$\displaystyle =\left[\frac{x}{2}\sqrt{4-x^{2}}+2\sin^{-1}\left(\frac{x}{2}\right)\right]_{0}^{2}-\left[2x-\frac{x^{2}}{2}\right]_{0}^{2}$
$\displaystyle =\left[0+2\sin^{-1}(1)-0\right]-\left[4-2\right]$
$\displaystyle =2\cdot\frac{\pi}{2}-2=\pi-2$
$\displaystyle \therefore \text{Area of region is }(\pi-2)\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 10: } \text{Find the area bounded by the curves } y=2x-x^{2} \text{ and the line } \\ y=x. \hspace{2.0cm} \text{ ISC 2013}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, curves are }y=2x-x^{2}\quad ...(i)$
$\displaystyle \text{and }y=x\quad ...(ii)$
$\displaystyle \text{From Eqs. (i) and (ii),}$
$\displaystyle x=2x-x^{2}$
$\displaystyle \Rightarrow x^{2}=x$
$\displaystyle \Rightarrow x(x-1)=0$
$\displaystyle \Rightarrow x=0,1$
$\displaystyle \text{When }x=0,\ y=0$
$\displaystyle \text{When }x=1,\ y=1$
$\displaystyle \therefore \text{Points of intersection are } \\ O(0,0)\text{ and }A(1,1)$
$\displaystyle \text{Required area}=\int_{0}^{1}\left[y_{\text{parabola}}-y_{\text{line}}\right]dx$
$\displaystyle =\int_{0}^{1}(2x-x^{2}-x)\,dx$
$\displaystyle =\int_{0}^{1}(x-x^{2})\,dx$
$\displaystyle =\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}$
$\displaystyle =\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\text{ sq unit}$
$\\$

$\displaystyle \textbf{Question 11: } \text{Find the area of the region bounded by the curves } \\ x=4y-y^{2} \text{ and the Y-axis.} \hspace{2.0cm} \text{ ISC 2012}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, curve is }x=4y-y^{2}$
$\displaystyle \Rightarrow x=-(y^{2}-4y)$
$\displaystyle \Rightarrow x=-(y^{2}-4y+4)+4$
$\displaystyle \Rightarrow (y-2)^{2}=4-x$
$\displaystyle \text{Let }y-2=Y\text{ and }x-4=X$
$\displaystyle \Rightarrow Y^{2}=-X$
$\displaystyle \text{This represents a parabola whose} \\ \text{axis is parallel to }X\text{-axis}$
$\displaystyle \text{Vertex is }(4,2)$
$\displaystyle \text{At }X\text{-axis, }y=0\Rightarrow x=0$
$\displaystyle \text{At }Y\text{-axis, }x=0\Rightarrow y^{2}-4y=0$
$\displaystyle \Rightarrow y(y-4)=0$
$\displaystyle \Rightarrow y=0,4$
$\displaystyle \text{Area bounded between parabola and }Y\text{-axis}$
$\displaystyle =\int_{0}^{4}x\,dy=\int_{0}^{4}(4y-y^{2})\,dy$
$\displaystyle =\left[\frac{4y^{2}}{2}-\frac{y^{3}}{3}\right]_{0}^{4}$
$\displaystyle =\left[2y^{2}-\frac{y^{3}}{3}\right]_{0}^{4}$
$\displaystyle =32-\frac{64}{3}=\frac{32}{3}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 12: } \text{Draw a rough sketch of the curve } y=(x-1)^{2} \text{ and } \\ y=|x-1|. \text{ Hence, find the area of the region bounded} \text{ by these curves.} \hspace{2.0cm} \text{ ISC 2011}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, curves are }y=(x-1)^{2}\quad ...(i)$ $\displaystyle \text{and }y=|x-1|=\begin{cases}x-1,&x\geq1\\-x+1,&x<1\end{cases}\quad ...(ii)$
$\displaystyle \text{From }y=(x-1)^{2}\text{ and }y=x-1,$
$\displaystyle (x-1)=(x-1)^{2}$
$\displaystyle \Rightarrow (x-1)^{2}-(x-1)=0$
$\displaystyle \Rightarrow (x-1)(x-2)=0$
$\displaystyle \Rightarrow x=1,2$
$\displaystyle \text{Thus, points of intersection are }P(1,0)\text{ and }Q(2,1)$
$\displaystyle \text{From }y=(x-1)^{2}\text{ and }y=-x+1,$
$\displaystyle (x-1)^{2}=-x+1$
$\displaystyle \Rightarrow x^{2}-2x+1=-x+1$
$\displaystyle \Rightarrow x^{2}-x=0$
$\displaystyle \Rightarrow x(x-1)=0$
$\displaystyle \Rightarrow x=0,1$
$\displaystyle \text{Thus, points of intersection are }R(0,1)\text{ and }P(1,0)$
$\displaystyle \therefore \text{Required area}=\int_{0}^{1}\left[-x+1-(x-1)^{2}\right]dx+\int_{1}^{2}\left[(x-1)-(x-1)^{2}\right]dx$
$\displaystyle =\int_{0}^{1}(x-x^{2})dx+\int_{1}^{2}(-x^{2}+3x-2)dx$
$\displaystyle =\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}+\left[-\frac{x^{3}}{3}+\frac{3x^{2}}{2}-2x\right]_{1}^{2}$
$\displaystyle =\left(\frac{1}{2}-\frac{1}{3}\right)+\left[\left(-\frac{8}{3}+6-4\right)-\left(-\frac{1}{3}+\frac{3}{2}-2\right)\right]$
$\displaystyle =\frac{1}{6}+\left[-\frac{2}{3}-\left(-\frac{5}{6}\right)\right]$
$\displaystyle =\frac{1}{6}+\frac{1}{6}=\frac{1}{3}\text{ sq unit}$
$\\$

$\displaystyle \textbf{Question 13: } \text{Draw a rough sketch of the curve } y=x^{2}-5x+6 \text{ and find the} \\ \text{area bounded by the curve and the X-axis.} \hspace{2.0cm} \text{ISC 2010}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, curve is }y=x^{2}-5x+6$
$\displaystyle =x^{2}-3x-2x+6$
$\displaystyle =x(x-3)-2(x-3)$
$\displaystyle =(x-3)(x-2)$
$\displaystyle \text{At }X\text{-axis, }y=0$
$\displaystyle \Rightarrow (x-3)(x-2)=0$
$\displaystyle \Rightarrow x=3\text{ or }x=2$
$\displaystyle \therefore \text{Required area}=\left|\int_{2}^{3}y\,dx\right|$
$\displaystyle =\left|\int_{2}^{3}(x^{2}-5x+6)dx\right|$
$\displaystyle =\left|\left[\frac{x^{3}}{3}-\frac{5x^{2}}{2}+6x\right]_{2}^{3}\right|$
$\displaystyle =\left|\left(9-\frac{45}{2}+18\right)-\left(\frac{8}{3}-10+12\right)\right|$
$\displaystyle =\left|\left(27-\frac{45}{2}\right)-\left(\frac{8}{3}+2\right)\right|$
$\displaystyle =\left|\frac{9}{2}-\frac{14}{3}\right|=\left|\frac{27-28}{6}\right|=\frac{1}{6}\text{ sq unit}$
$\\$

$\displaystyle \textbf{Question 14: } \text{Draw a rough sketch of the curve } x^{2}+y=9 \text{ and find the area} \\ \text{enclosed by the curve, the X-axis} \text{ and the lines } x+1=0 \text{ and } x-2=0. \hspace{2.0cm} \text{ ISC 2009}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, curves are }x^{2}+y=9\text{ and }x-2=0$ $\displaystyle \Rightarrow y=9-x^{2}$
$\displaystyle \text{This represents a parabola having vertex }(0,9)\text{ and opening downwards}$
$\displaystyle \therefore \text{Required area}=\int_{-1}^{2}y\,dx=\int_{-1}^{2}(9-x^{2})dx$
$\displaystyle =\left[9x-\frac{x^{3}}{3}\right]_{-1}^{2}$
$\displaystyle =\left(18-\frac{8}{3}\right)-\left(-9+\frac{1}{3}\right)$
$\displaystyle =18-\frac{8}{3}+9-\frac{1}{3}=27-3=24\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 15: } \text{Draw a rough sketch of the curve } y^{2}+1=x, \ x \leq 2. \\ \text{ Find the area enclosed by the curve and the line } x=2. \hspace{2.0cm} \text{ ISC 2008}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, curves are }y^{2}+1=x,\ x\leq2\quad ...(i)$ $\displaystyle \text{and }x=2\quad ...(ii)$
$\displaystyle \text{Now, }y^{2}=x-1$
$\displaystyle \Rightarrow y=\pm\sqrt{x-1}$
$\displaystyle \text{which is defined only when }x-1\geq0\Rightarrow x\geq1$
$\displaystyle \text{The curve }y^{2}=x-1\text{ is a right-open parabola with vertex }(1,0)$
$\displaystyle \therefore \text{Required area}=2\int_{1}^{2}y\,dx=2\int_{1}^{2}\sqrt{x-1}\,dx$
$\displaystyle \text{Put }x-1=t^{2}\Rightarrow dx=2t\,dt$
$\displaystyle \text{When }x=1,\ t=0;\quad \text{when }x=2,\ t=1$
$\displaystyle \therefore \text{Required area}=2\int_{0}^{1}t\cdot2t\,dt$
$\displaystyle =4\int_{0}^{1}t^{2}dt=4\left[\frac{t^{3}}{3}\right]_{0}^{1}=\frac{4}{3}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 16: } \text{Sketch and shade the area of the region lying in first quadrant and} \\ \text{bounded by } y=9x^{2}, \ x=0, \ y=1 \text{ and } y=4. \text{Find the area of the shaded region.} \hspace{2.0cm} \text{ISC 2004}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, curve is }y=9x^{2}$ $\displaystyle \Rightarrow x^{2}=\frac{y}{9},\text{ which represents a parabola}$
$\displaystyle \text{At }x=0,\ y=0$
$\displaystyle \text{At }y=1,\ x=\pm\sqrt{\frac{1}{9}}=\pm\frac{1}{3}$
$\displaystyle \text{At }y=4,\ x=\pm\sqrt{\frac{4}{9}}=\pm\frac{2}{3}$
$\displaystyle \therefore \text{Required area}=\text{Area of shaded region }(ABCDA)$
$\displaystyle =\int_{1}^{4}x\,dy\quad[\text{for the right branch of parabola}]$
$\displaystyle =\int_{1}^{4}\frac{\sqrt{y}}{3}\,dy$
$\displaystyle =\left[\frac{y^{3/2}}{3\cdot\frac{3}{2}}\right]_{1}^{4}$
$\displaystyle =\frac{2}{9}\left(4^{3/2}-1\right)$
$\displaystyle =\frac{2}{9}(8-1)=\frac{14}{9}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 17: } \text{Show that the area enclosed between the X-axis and the curve } \\ a^{2}y=x^{2}(x+a) \text{ is } \frac{a^{2}}{12}. \hspace{2.0cm} \text{ISC 2003}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, curve is }a^{2}y=x^{2}(x+a)\quad ...(i)$
$\displaystyle \text{At }X\text{-axis, }y=0$
$\displaystyle \Rightarrow x^{2}(x+a)=0$
$\displaystyle \Rightarrow x=0\text{ or }x=-a$
$\displaystyle \therefore \text{Required area}=\int_{-a}^{0}y\,dx$
$\displaystyle =\int_{-a}^{0}\frac{x^{2}(x+a)}{a^{2}}\,dx$
$\displaystyle =\int_{-a}^{0}\frac{x^{3}+ax^{2}}{a^{2}}\,dx$
$\displaystyle =\frac{1}{a^{2}}\left[\frac{x^{4}}{4}+\frac{ax^{3}}{3}\right]_{-a}^{0}$
$\displaystyle =\frac{1}{a^{2}}\left[0-\left(\frac{a^{4}}{4}-\frac{a^{4}}{3}\right)\right]$
$\displaystyle =\frac{1}{a^{2}}\left(\frac{a^{4}}{12}\right)$
$\displaystyle =\frac{a^{2}}{12}\text{ sq units}$
$\\$