Question

Using the geometrical meaning of the integral, calculate
${\int }_0^2\sqrt{2x-x^2}dx$

Answer 1

$I={\int }_0^2\sqrt{2x-x^2}dx$ $f\left(x\right)=2x-x^2$

$f(2a-x)=2(2a-x)-{(2-x)}^2$

$f(2-x)=f^1\left(x\right)$

$I=2{\int }_0^1\sqrt{2x-x^2}dx$

$=2{\int }_0^1\sqrt{-(x^2-2x)}dx=2{\int }_0^1\sqrt{-({(x-1)}^2+1)}dx$

$=2{\int }_0^1\sqrt{1-{(x-1)}^2}dx$

$=2{[\frac{1}{2}(x-1)\sqrt{1-{(x-1)}^2}+\frac{1}{2}{sin}^{-1}x-1]}_0^1+e$

$=0-(-1+0)=1$