Question

Evaluate : ${\int }_{-1}^0(e^{3x}+7x-5)dx$.

Answer 1

${\int }_{-1}^0(e^{3x}+7x-5)dx$

$={\int }_{-1}^0{e}^{3x}dx+7{\int }_{-1}^{0}xdx-5{\int }_{-1}^{0}dx$

${\left[\frac{e^{3x}}{3}\right]}_{-1}^0+{\left[7\frac{x^2}{2}\right]}_{-1}^0-5{\left[x\right]}_{-1}^0$

$=\frac{e^{-3\ast 0}}{3}-\frac{e^{-3}}{3}+7\frac{x\ast 0}{2}-7\frac{{(-1)}^2}{2}-5(0-(-1))$

$=\frac{1}{3}-\frac{1}{{3e}^3}-\frac{7}{2}-5$

$=\frac{2-21-30}{6}-\frac{1}{3e^3}$

$=-\frac{49}{6}-\frac{1}{{3e}^3}$

$=\frac{-98e^3-2}{6e^3}$