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 $x=a$.
$$\begin{aligned} &\text { Given equation of the curve is } y=\sqrt{a^2-x^2} \text {. }\\ &\Rightarrow \quad y^2=a^2-x^2 \Rightarrow y^2+x^2=a^2 \end{aligned}$$
$$\begin{aligned} &\therefore \text { Required area of shaded region, }\\ &\begin{aligned} A & =\int_0^a \sqrt{a^2-x^2} d x \\ & =\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]_0^a \\ & =\left[0+\frac{a^2}{2} \sin ^{-1}(1)-0-\frac{a^2}{2} \sin ^{-1} 0\right] \\ & =\frac{a^2}{2} \cdot \frac{\pi}{2}=\frac{\pi a^2}{4} \text { sq units } \end{aligned} \end{aligned}$$
Find the area of the region bounded by $y=\sqrt{x}$ and $y=x$.
$$\begin{aligned} & \text { Given equation of curves are } \quad y=\sqrt{x} \text { and } y=x \text {. } \\ & \Rightarrow \quad x=\sqrt{x} \Rightarrow x^2=x \\ & \Rightarrow \quad x^2-x=0 \Rightarrow x(x-1)=0 \\ & \Rightarrow \quad x=0,1 \end{aligned}$$
$$\begin{aligned} &\therefore \text { Required area of shaded region, }\\ &\begin{aligned} A & =\int_0^1(\sqrt{x}) d x-\int_0^1 x d x \\ & =\left[2 \cdot \frac{x^{3 / 2}}{3}\right]_0^1-\left[\frac{x^2}{2}\right]_0^1 \\ & =\frac{2}{3} \cdot 1-\frac{1}{2}=\frac{2}{3}-\frac{1}{2}=\frac{1}{6} \text { sq units } \end{aligned} \end{aligned}$$
Find the area enclosed by the curve $y=-x^2$ and the straight line $x+y+2=0$.
We have, $y=-x^2$ and $x+y+2=0$
$$\begin{array}{lrl} \Rightarrow & -x-2 & =-x^2 \Rightarrow x^2-x-2=0 \\ \Rightarrow & x^2+x-2 x-2 & =0 \Rightarrow x(x+1)-2(x+1)=0 \\ \Rightarrow & (x-2)(x+1) & =0 \Rightarrow x=2,-1 \end{array}$$
$$ \begin{aligned} \therefore \quad \text { Area of shaded region, }A & =\left|\int_{-1}^2\left(-x-2+x^2\right) d x\right|=\left|\int_{-1}^2\left(x^2-x-2\right) d x\right| \\ & =\left|\left[\frac{x^3}{3}-\frac{x^2}{2}-2 x\right]_{-1}^2\right|=\left|\left[\frac{8}{3}-\frac{4}{2}-4+\frac{1}{3}+\frac{1}{2}-2\right]\right| \\ & =\left|\frac{16-12-24+2+3-12}{6}\right|=\left|-\frac{27}{6}\right|=\frac{9}{2} \text { sq units } \end{aligned}$$