Question

If $f\left(x\right)=\left|x\right|+1,-1\le x<0$
$1+{\left|x\right|}^2,0\le x\le 1$
then ${\int }_{-1}^{1}f\left(x\right)dx$ is equal to

• A. $-\frac{1}{6}$
• B. $\frac{17}{6}$
• C. $-\frac{17}{6}$
• D. none of these

Answer 1

Answer: (B)

$\frac{17}{6}$

$I={\int }_{-1}^{1}f\left(x\right)dx={\int }_{-1}^{0}f\left(x\right)dx+{\int }_0^{1}f\left(x\right)dx$
$I={\int }_{-1}^0(-x+1)dx+{\int }_0^1(1+x^2)dx={[-\frac{x^2}{2}+x]}_{-1}^0+{[x+\frac{x^3}{3}]}_0^1=(0+\frac{1}{2}+1)+(1+\frac{1}{3}-0)=\frac{3}{2}+\frac{4}{3}=\frac{17}{6}$