Question

${\int }_1^3(x^2+3x+e^x)dx$.

Answer 1

${\int }_1^3(x^2+3x+e^x)dx$

$={[\frac{x^3}{3}+\frac{3x^2}{2}+e^x]}_1^3$

$=9+\frac{27}{2}+e^3-\frac{1}{3}-\frac{3}{2}-e^1$

$=9+12-\frac{1}{3}+e^2(e-1)$

$=\frac{62}{3}+e^2(e-1)$

Hence, the answer is $\frac{62}{3}+e^2(e-1).$