Question

Evaluate: ${\int }_0^2|2x-x^2|dx$

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{4}{3}$
• D. $1$

Answer 1

The correct option is C $\frac{4}{3}$

$-x^2+2x=0$ maximum minimum at $x=\frac{-2}{-1\times 2}=1$

$x=\frac{-2\pm \sqrt{2^2-4(-1)\left(0\right)}}{2(-1)}=\frac{-2\pm 2}{-2}=2,0$

the function

$2x-x^2$ is positive in 0 to 2

$\therefore |2x-x^2|=2x-x^2$ in 0 to 2

${\int }_0^2|2x-x^2|={\int }_0^2{2x-x}^2$

$={\int }_0^{2}2x-x^2={\int }_0^{2}2x-{\int }_0^2{x}^2=2\times \frac{x^2}{2}-\frac{x^3}{3}=x^2-\frac{x^3}{3}=\left(2^2\right)-(\frac{2^3}{3}-0)=\frac{4}{3}$