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 $x=8$.
$$\begin{aligned} \text{We have,}\quad 2 y & =5 x+7 \\ \Rightarrow \quad y & =\frac{5 x}{2}+\frac{7}{2} \end{aligned}$$
$$\begin{aligned} \therefore \text { Area of shaded region } & =\frac{1}{2} \int_2^8(5 x+7) d x=\frac{1}{2}\left[5 \cdot \frac{x^2}{2}+7 x\right]_2^8 \\ & =\frac{1}{2}[5 \cdot 32+7 \cdot 8-10-14]=\frac{1}{2}[160+56-24] \\ & =\frac{192}{2}=96 \text { sq units } \end{aligned}$$
Draw a rough sketch of the curve $y=\sqrt{x-1}$ in the interval $[1,5]$. Find the area under the curve and between the lines $x=1$ and $x=5$.
$$\begin{aligned} &\text { Given equation of the curve is } y=\sqrt{x-1} \text {. }\\ &\Rightarrow \quad y^2=x-1 \end{aligned}$$
$$\begin{aligned} \therefore \text { Area of shaded region, }A & =\int_1^5(x-1)^{1 / 2} d x=\left[\frac{2 \cdot(x-1)^{3 / 2}}{3}\right]_1^5 \\ & =\left[\frac{2}{3} \cdot(5-1)^{3 / 2}-0\right]=\frac{16}{3} \text { sq units } \end{aligned}$$
Determine the area under the curve $y=\sqrt{a^2-x^2}$ included between the lines $x=0$ and $x=a$.
$$\begin{aligned} &\text { Given equation of the curve is } y=\sqrt{a^2-x^2} \text {. }\\ &\Rightarrow \quad y^2=a^2-x^2 \Rightarrow y^2+x^2=a^2 \end{aligned}$$
$$\begin{aligned} &\therefore \text { Required area of shaded region, }\\ &\begin{aligned} A & =\int_0^a \sqrt{a^2-x^2} d x \\ & =\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]_0^a \\ & =\left[0+\frac{a^2}{2} \sin ^{-1}(1)-0-\frac{a^2}{2} \sin ^{-1} 0\right] \\ & =\frac{a^2}{2} \cdot \frac{\pi}{2}=\frac{\pi a^2}{4} \text { sq units } \end{aligned} \end{aligned}$$