Question

Let $f\left(x\right)=min\{x,1-x\}$ for all $xϵR$. Then the value of ${\int }_0^{2}f\left(x\right)dx$ is

• A. $-\frac{1}{4}$
• B. $\frac{1}{4}$
• C. $2$
• D. $0$

Answer 1

Answer: (A)

$-\frac{1}{4}$

${\int }_0^{2}f\left(x\right)dx={\int }_0^{2}min\{x,1-x\}dx$
$={\int }_0^{\frac{1}{2}}xdx+{\int }_{\frac{1}{2}}^2(1-x)dx$
$={\left[\frac{x^2}{2}\right]}_0^{\frac{1}{2}}+{[x-\frac{x^2}{2}]}_{\frac{1}{2}}^2=-\frac{1}{4}$