Using integration, find the area of triangle ABC, whose vertices are A(2, 5), B(4, 7)
and C(6, 2).

$\mathbf{OR}$

Find the area of the region lying above x-axis and included between the circle x²+y²=8x and inside of the parabola y²=4x.

Draw the graph from the given information,

Notice that the vertices of the given triangle are A(2, 5), B(4, 7) and C(6, 2).

Equation of AB

$$\begin{gathered} \mathrm{y}-5=\frac{7-5}{4-2}(\mathrm{x}-2) \\ \Rightarrow \mathrm{y}-5=\mathrm{x}-2 \\ \Rightarrow \mathrm{y}=\mathrm{x}+3 \end{gathered}$$

Let say ${\mathrm{y}}_1=\mathrm{x}+3$

Equation of BC :

$$\begin{gathered} \mathrm{y}-7=\frac{2-7}{6-4}(\mathrm{x}-4) \\ \Rightarrow \mathrm{y}=\frac{-5}{2}(\mathrm{x}-4)+7=\frac{-5}{2}\mathrm{x}+17 \end{gathered}$$

Let's say ${\mathrm{y}}_2=-\frac{5}{2}\mathrm{x}+17$

Equation of AC :

$$\begin{gathered} \mathrm{y}-5=\frac{2-5}{6-2}(\mathrm{x}-2) \\ \Rightarrow \mathrm{y}=\frac{-3}{4}(\mathrm{x}-2)+5=\frac{-3}{4}\mathrm{x}+\frac{13}{2} \end{gathered}$$

Let say ${\mathrm{y}}_3=\frac{3}{4}\mathrm{x}+\frac{13}{2}$

$$\begin{gathered} \mathrm{ar} (∆ \mathrm{ABC})={\int }_2^4 {\mathrm{y}}_1 \mathrm{dx}+{\int }_4^6 {\mathrm{y}}_2 \mathrm{dx}-{\int }_2^6 {\mathrm{y}}_3 \mathrm{dx} \\ ={\int }_2^4 (\mathrm{x}+3) \mathrm{dx}+{\int }_4^6 \left(\frac{-5}{2}\mathrm{x}+17\right) \mathrm{dx}-{\int }_2^{6 }\left(\frac{-3}{4}\mathrm{x}+\frac{13}{2}\right) \mathrm{dx} \\ ={\left[\frac{{\mathrm{x}}^2}{2}+3\mathrm{x}\right]}_2^4+{\left[\frac{-5{\mathrm{x}}^2}{4}+17\mathrm{x}\right]}_4^6-{\left[\frac{-3{\mathrm{x}}^2}{8}+\frac{13\mathrm{x}}{2}\right]}_2^6 \\ =\left[\frac{16}{2}+12-\frac{4}{2}-6\right]+\left[\frac{-180}{4}+102+\frac{80}{4}-68\right]-\left[\frac{-108}{8}+\frac{78}{2}+\frac{12}{8}-\frac{26}{2}\right] \\ =12+9+14 \\ 35 \mathrm{sq}. \mathrm{units}. \end{gathered}$$

Therefore, the area of the triangle ABC is 35 unit².

                                         OR

 Draw the graph from the given information,

Given equations are,

x²+y²=8x……(1)

y²=4x………..(2)

Clearly the equation x²+y²=8x is a circle with centre (4, 0) and has a radius 4. Also y²=4x is a parabola with vertex at origin and the axis along the x-axis opening in the positive direction.

To find the intersecting points of the curves, we solve both the equation.

$$\begin{gathered} \therefore {\mathrm{x}}^2+4\mathrm{x}=8\mathrm{x} \\ \Rightarrow {\mathrm{x}}^2-4\mathrm{x}=0 \\ \Rightarrow \mathrm{x}(\mathrm{x}-4)=0 \\ \Rightarrow \mathrm{x}=0 \mathrm{and} \mathrm{x}=4 \end{gathered}$$

When x=0, y=0

When x=4, y=$\pm$4

To approximate the area of the shaded region the length $|{\mathrm{y}}_2-{\mathrm{y}}_1|$ and the width =dx

$$\begin{gathered} \mathrm{A}={\int }_0^4 |{\mathrm{y}}_2-{\mathrm{y}}_1| \mathrm{dx} \\ ={\int }_0^4 ({\mathrm{y}}_2-{\mathrm{y}}_1) \mathrm{dx} \left[\because {\mathrm{y}}_2>{\mathrm{y}}_1 \therefore |{\mathrm{y}}_2-{\mathrm{y}}_1|={\mathrm{y}}_2-{\mathrm{y}}_1\right] \\ ={\int }_0^4\left[\sqrt{\left(16-{(\mathrm{x}-4)}^2\right)}-\sqrt{4\mathrm{x}}\right] \mathrm{dx} \left\{\therefore {\mathrm{y}}_2=\sqrt{16-{(\mathrm{x}-4)}^2} \mathrm{and} {\mathrm{y}}_1=2\sqrt{\mathrm{x}}\right\} \\ ={\int }_0^4 \sqrt{16-{(\mathrm{x}-4)}^2} \mathrm{dx}-{\int }_0^4 \sqrt{4\mathrm{x}} \mathrm{dx} \\ ={{\left[\frac{(\mathrm{x}-4)}{2} \sqrt{16-{(\mathrm{x}-4)}^2}+\frac{16}{2}{\sin }^{-1} \left(\frac{\mathrm{x}-4}{4}\right)\right]}^4}_0-{{\left[\frac{4{\mathrm{x}}^{{\displaystyle \frac{3}{2}}}}{3}\right]}^4}_0 \\ =\left[0+0-0-8{\sin }^{-1} \left(\frac{-4}{4}\right)\right]-\frac{4}{3}\times 4^{\frac{3}{2}} \\ =\frac{8\mathrm{\pi }}{2}-\frac{32}{3} \\ =4\mathrm{\pi }-\frac{32}{3} \end{gathered}$$Therefore, the area of the region lying above x-axis and included between the circle x²+y²=8x and inside of the parabola y²=4x is $4\mathrm{\pi }-\frac{32}{3}$ unit².