Answer
$\int_{1}^2(8x^3+3x^2)dx=37$
Work Step by Step
$\int_{1}^2(8x^3+3x^2)dx$
$=[\frac{8x^4}{4}+\frac{3x^3}{3}]^2_{1}$
$=[2x^4+x^3]^2_{1}$
Sub in the bounds and subtract the lower bound (1) from the upper bound (2).
$=[2(2)^4+(2)^3]-[2(1)^4+(1)^3]$
$=40-3$
$=37$