Question

Evaluate the integral

${\int }_8^9\sqrt{\frac{x-8}{9-x}}dx$

• A. $\pi$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{6\sqrt{2}}$

Answer 1

The correct option is C $\frac{\pi }{2}$

$I={\int }_8^9\sqrt{\frac{x-8}{9-x}}dx$ ---(1)

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_8^9\sqrt{\frac{9-x}{x-8}}dx$ ---(2)

Adding $1$ and $2$, we get

$2I={\int }_8^9\frac{x-8+9-x}{\left(\sqrt{x-8}\right)\left(\sqrt{9-x}\right)}$

$2I={\int }_8^9\frac{1}{\sqrt{(x-8)(9-x)}}dx={\int }_8^9\frac{dx}{\sqrt{(0.5{)}^2}-(x-8.5{)}^2}$

$=si{n}^{-1}\left(\frac{x-8.5}{0.5}\right){\int }_8^9$

$2I=si{n}^{-1}\left(1\right)-si{n}^{-1}(-1)$

$I=\frac{\pi }{2}$