Question

Evaluate the following the sums:
${\int }_0^2(x^2+1)dx$

Answer 1

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

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

$=\frac{1}{3}(2^3-0)+(2-0)$

$=\frac{8}{3}+2=\frac{8+6}{3}=\frac{14}{3}$