Question

Evaluate the integral

${\int }_1^2\sqrt{(x-1)(2-x)}dx$

• A. $\frac{\pi }{8}$
• B. $\frac{\pi }{4}$
• C. $\frac{1}{8}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{\pi }{8}$

${\int }_1^2\sqrt{-x^2+3x-2}dx$

$={\int }_1^2\sqrt{-x^2+3x-\frac{9}{4}+\frac{9}{4}-2}dx$

$={\int }_1^2\sqrt{-{(x-\frac{3}{2})}^2+{\left(\frac{1}{2}\right)}^2}dx={\int }_1^2\sqrt{{\left(\frac{1}{2}\right)}^2-{(x-\frac{3}{2})}^2}dx$

$={(\frac{x}{2}\sqrt{(x-1)(2-x)+\frac{(\frac{1}{2}{)}^2}{2}}{\sin }^{-1}\left(\frac{x-\frac{3}{2}}{\frac{1}{2}}\right)+c)}_1^2$

$={(\frac{x}{2}\sqrt{(x-1)(2-x)}+\frac{1}{8}{\sin }^{-1}(2x-3)+c)}_1^2$

$=\frac{1}{8}{\sin }^{-1}\left(1\right)-\frac{1}{8}{\sin }^{-1}(-1)$

$=\frac{1}{8}(\frac{\pi }{2}+\frac{\pi }{8})$

$=\frac{\pi }{8}$