# Evaluate the following definite integrals \int_{0}^{1}(x^{4}+7e^{x}-3)dx

Evaluate the following definite integrals
${\int }_{0}^{1}\left({x}^{4}+7{e}^{x}-3\right)dx$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

cyhuddwyr9

Given that
${\int }_{0}^{1}\left({x}^{4}+7{e}^{x}-3\right)dx$

$=\frac{1}{5}+7{e}^{1}-3\left(1\right)-0-7{e}^{0}+3\left(0\right)$
$=\frac{1}{5}+7e\cdot -10$
$=7e-\frac{49}{5}$
$\therefore {\int }_{0}^{1}\left({x}^{4}+7{e}^{x}-3\right)dx=7e-\frac{49}{5}$