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

$\displaystyle \textbf{Question 1: } \text{If the area enclosed between the curves } y=ax^{2} \text{ and } x=ay^{2}, \\ a>0, \text{ is } 27 \text{ sq unit, find the value of } a.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have two parabolas, }y=ax^{2}\quad ...(i)$
$\displaystyle \text{and }x=ay^{2}\quad ...(ii)$
$\displaystyle \text{Putting }y=ax^{2}\text{ in Eq. (ii), we get}$
$\displaystyle x=a(ax^{2})^{2}$
$\displaystyle \Rightarrow x=a^{3}x^{4}$
$\displaystyle \Rightarrow x(1-a^{3}x^{3})=0$
$\displaystyle \Rightarrow x=0\text{ or }1-a^{3}x^{3}=0$
$\displaystyle \Rightarrow x=0\text{ or }x=\frac{1}{a}$
$\displaystyle \text{When }x=0,\ y=0$
$\displaystyle \text{When }x=\frac{1}{a},\ y=\frac{1}{a}$
$\displaystyle \therefore \text{Points of intersection are }O(0,0)\text{ and }B\left(\frac{1}{a},\frac{1}{a}\right)$
$\displaystyle \text{Required area}=\int_{0}^{1/a}\left(\sqrt{\frac{x}{a}}-ax^{2}\right)dx=27$
$\displaystyle \Rightarrow \left[\frac{1}{\sqrt{a}}\cdot\frac{x^{3/2}}{3/2}-\frac{ax^{3}}{3}\right]_{0}^{1/a}=27$
$\displaystyle \Rightarrow \frac{2}{3a^{2}}-\frac{1}{3a^{2}}=27$
$\displaystyle \Rightarrow \frac{1}{3a^{2}}=27$
$\displaystyle \Rightarrow 3a^{2}=\frac{1}{27}$
$\displaystyle \Rightarrow a^{2}=\frac{1}{81}$
$\displaystyle \Rightarrow a=\frac{1}{9}\quad[\because a>0]$
$\\$

$\displaystyle \textbf{Question 2: } \text{If the area is bounded by the parabola } y^{2}=16x \text{ and the line } \\ y=4mx \text{ is } \frac{1}{12} \text{ sq unit,} \text{then using integration, find the value of } m \ (m>0).$
$\displaystyle \text{Answer:}$
$\displaystyle\text{Let }y^{2}=16x \quad ...(i)$
$\displaystyle \text{and }y=4mx \quad ...(ii)$
$\displaystyle \text{From Eqs. (i) and (ii), we get}$
$\displaystyle 16m^{2}x^{2}=16x$
$\displaystyle \Rightarrow x=\frac{1}{m^{2}}\text{ or }x=0$
$\displaystyle \text{According to the question,} \\ \text{area bounded by the curves is}$
$\displaystyle \int_{0}^{1/m^{2}}(y_{1}-y_{2})\,dx=\frac{1}{12}$
$\displaystyle \Rightarrow \int_{0}^{1/m^{2}}\left(\sqrt{16x}-4mx\right)dx=\frac{1}{12}$
$\displaystyle \Rightarrow \left[\frac{4x^{3/2}}{3/2}-\frac{4mx^{2}}{2}\right]_{0}^{1/m^{2}}=\frac{1}{12}$
$\displaystyle \Rightarrow \left[\frac{8x^{3/2}}{3}-2mx^{2}\right]_{0}^{1/m^{2}}=\frac{1}{12}$
$\displaystyle \Rightarrow \frac{8}{3}\left(\frac{1}{m^{2}}\right)^{3/2}-2m\left(\frac{1}{m^{2}}\right)^{2}=\frac{1}{12}$
$\displaystyle \Rightarrow \frac{8}{3m^{3}}-\frac{2}{m^{3}}=\frac{1}{12}$
$\displaystyle \Rightarrow \frac{8-6}{3m^{3}}=\frac{1}{12}$
$\displaystyle \Rightarrow m^{3}=8$
$\displaystyle \therefore m=2$
$\\$

$\displaystyle \textbf{Question 3: } \text{Find the area of the region bounded by the curves } \\ y=x^{2}+2, \ y=x, \ x=0 \text{ and } x=3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, curve }y=x^{2}+2$
$\displaystyle \text{i.e. }y-2=x^{2}\text{ represents a parabola with vertex }(0,2)$
$\displaystyle \text{and symmetric about }Y\text{-axis}$ $\displaystyle \text{The area bounded by }y=x^{2}+2,\ y=x,\ x=0\text{ and }x=3$
$\displaystyle \text{is represented by the shaded region}$
$\displaystyle \therefore \text{Required area}=\text{area }(OABDO)-\text{area }(OCDO)$
$\displaystyle =\int_{0}^{3}(x^{2}+2)\,dx-\int_{0}^{3}x\,dx$
$\displaystyle =\left[\frac{x^{3}}{3}+2x\right]_{0}^{3}-\left[\frac{x^{2}}{2}\right]_{0}^{3}$
$\displaystyle =\left[\frac{3^{3}}{3}+6-0\right]-\left[\frac{3^{2}}{2}-0\right]$
$\displaystyle =9+6-\frac{9}{2}=\frac{21}{2}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 4: } \text{Using integration, find the area of the region bounded between the line } \\ x=4 \text{ and the parabola } y^{2}=16x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }y^{2}=16x \quad ...(i)$
$\displaystyle x=4 \quad ...(ii)$ $\displaystyle \therefore \text{Required area}=2\int_{0}^{4}y\,dx=2\int_{0}^{4}\sqrt{16x}\,dx$
$\displaystyle =8\int_{0}^{4}\sqrt{x}\,dx$
$\displaystyle =8\left[\frac{2}{3}x^{3/2}\right]_{0}^{4}$
$\displaystyle =\frac{16}{3}\left[4^{3/2}-0\right]$
$\displaystyle =\frac{16}{3}\times8=\frac{128}{3}\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 5: } \text{Sketch the graph of } y=|x+4|. \text{ Using integration, find the area of the } \\ \text{region bounded by the curve } y=|x+4| \text{ and } x=-6 \text{ and } x=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }y=|x+4|\Rightarrow y=\begin{cases}-(x+4),&x<-4\\x+4,&x\geq -4\end{cases}$ $\displaystyle \text{Area of shaded region}=\int_{-6}^{-4}y_{1}\,dx+\int_{-4}^{0}y_{2}\,dx$
$\displaystyle \text{where }y_{1}=-x-4\text{ and }y_{2}=x+4$
$\displaystyle \therefore A=\int_{-6}^{-4}(-x-4)\,dx+\int_{-4}^{0}(x+4)\,dx$
$\displaystyle =-\left[\frac{x^{2}}{2}+4x\right]_{-6}^{-4}+\left[\frac{x^{2}}{2}+4x\right]_{-4}^{0}$
$\displaystyle =-\left[\left(\frac{16}{2}-16\right)-\left(\frac{36}{2}-24\right)\right]+\left[0-\left(\frac{16}{2}-16\right)\right]$
$\displaystyle =-\left[-8-(-6)\right]+[0-(-8)]$
$\displaystyle =2+8=10\text{ sq units}$
$\\$

$\displaystyle \textbf{Question 6: } \text{Using integration, find the area of the following region } \\ \{(x,y): \frac{x^{2}}{9}+\frac{y^{2}}{4} \leq 1 \leq \frac{x}{3}+\frac{y}{2}\}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, curve is }\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
$\displaystyle \text{It is an ellipse with centre }(0,0)$
$\displaystyle \text{and given line is }\frac{x}{3}+\frac{y}{2}=1$
$\displaystyle \text{For points of intersection, from the line, }\frac{x}{3}=1-\frac{y}{2}$
$\displaystyle \text{Substituting in the ellipse, we get}$
$\displaystyle \left(1-\frac{y}{2}\right)^{2}+\frac{y^{2}}{4}=1$
$\displaystyle \Rightarrow 1-y+\frac{y^{2}}{4}+\frac{y^{2}}{4}=1$
$\displaystyle \Rightarrow y^{2}-2y=0$
$\displaystyle \Rightarrow y=0,2$
$\displaystyle \text{When }y=0,\ \frac{x}{3}=1\Rightarrow x=3$
$\displaystyle \text{When }y=2,\ \frac{x}{3}+1=1\Rightarrow x=0$
$\displaystyle \text{Thus, points of intersection are }A(3,0)\text{ and }B(0,2)$
$\displaystyle \therefore \text{Required area}=\int_{0}^{3}\left[y_{\text{ellipse}}-y_{\text{line}}\right]dx$
$\displaystyle =\int_{0}^{3}2\sqrt{1-\frac{x^{2}}{9}}\,dx-\int_{0}^{3}2\left(1-\frac{x}{3}\right)dx$
$\displaystyle =\frac{2}{3}\int_{0}^{3}\sqrt{9-x^{2}}\,dx-\frac{2}{3}\int_{0}^{3}(3-x)dx$
$\displaystyle =\frac{2}{3}\left[\frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\sin^{-1}\frac{x}{3}\right]_{0}^{3}-\frac{2}{3}\left[3x-\frac{x^{2}}{2}\right]_{0}^{3}$
$\displaystyle =\frac{2}{3}\left(0+\frac{9}{2}\sin^{-1}(1)-0\right)-\frac{2}{3}\left(9-\frac{9}{2}\right)$
$\displaystyle =\frac{2}{3}\cdot\frac{9}{2}\cdot\frac{\pi}{2}-\frac{2}{3}\cdot\frac{9}{2}$
$\displaystyle =\frac{3\pi}{2}-3=3\left(\frac{\pi}{2}-1\right)\text{ sq units}$
$\\$