Question

Evaluate the following define integral as limit of sums :
${\int }_0^4(3x^2+2x+1)dx$

Answer 1

${\int }_0^4(3x^2+2x+1)dx$

$={[\frac{3x^3}{3}+\frac{2x^2}{2}+x]}_0^4$

$={[x^3+x^2+x]}_0^4$

$=(4^3+4^2+4)-\left(0\right)$

$=64+16+4=84$