Question

Evaluate

${\int }_a^b\sqrt{(x-a)(b-x)}dx,a>b$

Answer 1

$\begin{array}{c}{\int }_a^b\sqrt{\left(x-1\right)\left(b-x\right)}dx,a>b\\ {\int }_a^b\sqrt{-x^2+\left(a+b\right)x-ab}dx\\ {\int }_a^b\sqrt{-x^2+\left(a+b\right)x-{\left(\frac{a+b}{2}\right)}^2+{\left(\frac{a+b}{2}\right)}^2-ab}dx\\ {\int }_a^b\sqrt{-\left[x^2-\left(a+b\right)x+{\left(\frac{a+b}{2}\right)}^2+\frac{a^2+b^2+2ab-ab}{4}\right]}dx\\ {\int }_a^b\sqrt{-{\left(x-\frac{a+b}{2}\right)}^2+\frac{a^2+b^2-2ab}{4}}dx\\ {\int }_a^b\sqrt{-{\left(x-\left(\frac{a+b}{2}\right)\right)}^2+{\left(\frac{a-b}{2}\right)}^2}dx\\ {\left[\frac{x-\left(\frac{a+b}{2}\right)\sqrt{\left(x-1\right)\left(b-x\right)}+}{2}\frac{{\left(\frac{a-b}{2}\right)}^2{\sin }^{-1}}{2}x-\frac{\frac{\left(a-b\right)}{2}}{\frac{a-b}{2}}\right]}_a^b\\ {\left(\frac{a-b}{8}\right)}^2\left[{\sin }^{-1}\frac{b-\left(\frac{a+b}{2}\right)}{\frac{a-b}{2}}-{\sin }^{-1}\frac{a-\left(\frac{a+b}{2}\right)}{\frac{a-b}{2}}\right]\\ {\left(\frac{a-b}{8}\right)}^2\left[{\sin }^{-1}\left(-1\right)-{\sin }^{-1}\left(1\right)\right]\\ {\left(\frac{a-b}{8}\right)}^2\left[-2{\sin }^{-1}1\right]\\ -{\left(\frac{a-b}{8}\right)}^2\pi \end{array}$