Question
$\int_0^4\left(e^{2 x}+x\right) d x$ is equal to

Answer

$\text {Let } I=\int_0^4\left(e^{2 x}+x\right) d x=\left[\frac{e^{2 x}}{2}+\frac{x^2}{2}\right]_0^4$
$=\frac{e^8}{2}+\frac{16}{2}-\frac{e^0}{2}-0=\frac{e^8}{2}+\frac{16}{2}-\frac{1}{2}=\frac{e^8+15}{2}$

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