Question

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

Answer 1

The given integral is:

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

Using the property of definite integrals: ${\int }_a^b\left(f\right(x)+g(x\left)\right)dx={\int }_a^{b}f\left(x\right)dx+{\int }_a^{b}g\left(x\right)dx$, the above integral becomes:

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

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

Substituting the limits and adding up to get the result, we get:

$\frac{27-1}{3}+\frac{3}{2}(9-1)+(e^3-e)=e^3-e+\frac{62}{3}$