Find the area of the region enclosed by the parabola $x^2=y$ and the line $y=x+2$.
$$\begin{aligned} &\text { Sol. We have, } x^2=y \text { and } y=x+2\\ &\begin{array}{r} \Rightarrow & x^2 =x+2 \\ \Rightarrow & x^2-x-2 =0 \\ \Rightarrow & x^2-2 x+x-2 =0 \\ \Rightarrow & x(x-2)+1(x-2) =0 \\ \Rightarrow & (x+1)(x-2) =0 \\ \Rightarrow & x =-1,2 \end{array} \end{aligned}$$
$$\begin{aligned} \therefore \text { Required area of shaded region } & =\int_{-1}^2\left(x+2-x^2\right) d x=\left[\frac{x^2}{2}+2 x-\frac{x^3}{3}\right]_{-1}^2 \\ & =\left[\frac{4}{2}+4-\frac{8}{3}-\frac{1}{2}+2-\frac{1}{3}\right] \\ & =6+\frac{3}{2}-\frac{9}{3}=\frac{36+9-18}{6}=\frac{27}{6}=\frac{9}{2} \text { sq units } \end{aligned}$$
Find the area of the region bounded by line $x=2$ and parabola $y^2=8 x$.
We have, $y^2=8 x$ and $x=2$
$$\begin{aligned} \therefore \text { Area of shaded region, } A & =2 \int_0^2 \sqrt{8 x} d x=2 \cdot 2 \sqrt{2} \int_0^2 x^{1 / 2} d x \\ & =4 \cdot \sqrt{2} \cdot\left[2 \cdot \frac{x^{3 / 2}}{3}\right]_0^2=4 \sqrt{2}\left[\frac{2}{3} \cdot 2 \sqrt{2}-0\right] \\ & =\frac{32}{3} \text { sq units } \end{aligned}$$
Sketch the region $\left\{(x, 0): y=\sqrt{4-x^2}\right\}$ and $X$-axis. Find the area of the region using integration.
Given region is $\left\{(x, 0): y=\sqrt{4-x^2}\right\}$ and $X$-axis.
We have, $$y=\sqrt{4-x^2} \Rightarrow y^2=4-x^2 \Rightarrow x^2+y^2=4$$
$$\begin{aligned} & \therefore \quad \text { Area of shaded region, } A=\int_{-2}^2 \sqrt{4-x^2} d x=\int_{-2}^2 \sqrt{2^2-x^2} d x \\ & =\left[\frac{x}{2} \sqrt{2^2-x^2}+\frac{2^2}{2} \cdot \sin ^{-1} \frac{x}{2}\right]_{-2}^2 \\ & =\frac{2}{2} \cdot 0+2 \cdot \frac{\pi}{2}+\frac{2}{2} \cdot 0-2 \sin ^{-1}(-1)=2 \cdot \frac{\pi}{2}+2 \cdot \frac{\pi}{2} \\ & =2 \pi \text { sq units } \end{aligned}$$
Calculate the area under the curve $y=2 \sqrt{x}$ included between the lines $x=0$ and $x=1$.
We have, $y=2 \sqrt{x}, x=0$ and $x=1$
$$\begin{aligned} \therefore \quad \text { Area of shaded region, } A & =\int_0^1(2 \sqrt{x}) d x \\ & =2 \cdot\left[\frac{x^{3 / 2}}{3} \cdot 2\right]_0^1 \\ & =2\left(\frac{2}{3} \cdot 1-0\right)=\frac{4}{3} \text { sq units } \end{aligned}$$
Using integration, find the area of the region bounded by the line $2 y=5 x+7, X$-axis and the lines $x=2$ and $x=8$.
$$\begin{aligned} \text{We have,}\quad 2 y & =5 x+7 \\ \Rightarrow \quad y & =\frac{5 x}{2}+\frac{7}{2} \end{aligned}$$
$$\begin{aligned} \therefore \text { Area of shaded region } & =\frac{1}{2} \int_2^8(5 x+7) d x=\frac{1}{2}\left[5 \cdot \frac{x^2}{2}+7 x\right]_2^8 \\ & =\frac{1}{2}[5 \cdot 32+7 \cdot 8-10-14]=\frac{1}{2}[160+56-24] \\ & =\frac{192}{2}=96 \text { sq units } \end{aligned}$$