Question

Evaluate the integral

${\int }_{-1}^1|1+x^2|dx$

• A. $\frac{8}{3}$
• B. $1$
• C. $-\frac{1}{3}$
• D. $-\frac{4}{3}$

Answer 1

The correct option is A $\frac{8}{3}$

${\int }_{-1}^1(1+x^2)dx$

$1+x^2>0$

${\int }_{-1}^1(1+x^2)dx$

$1+x^2$ is even function

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

$=2[x+x\frac{3}{3}+c{]}_0^1$

$=2\left[\right(1+\frac{1}{3}+c)-(0+c\left)\right]$

$=\frac{8}{3}$

Thus ${\int }_{-1}^1(1+x^2)dx=\frac{8}{3}$