Question

# Evaluate the following definite integrals: int_0^1(x^4+7e^x-3)dx

Applications of integrals
Evaluate the following definite integrals:
$$\displaystyle{\int_{{0}}^{{1}}}{\left({x}^{{4}}+{7}{e}^{{x}}-{3}\right)}{\left.{d}{x}\right.}$$

2020-11-21
Step 1
To Evaluate the following definite integrals:
Step 2
Given That
$$\displaystyle{\int_{{0}}^{{1}}}{\left({x}^{{4}}+{7}{e}^{{x}}-{3}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle={{\left[\frac{{x}^{{5}}}{{5}}+{7}{e}^{{x}}-{3}{x}\right]}_{{0}}^{{1}}}{\left[\begin{array}{c} \int{x}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}^{{n}}+{1}}}{{{n}+{1}}}+{c}\\\int{e}^{{x}}{\left.{d}{x}\right.}={e}^{{x}}+{c}\end{array}\right]}$$
$$\displaystyle=\frac{{1}}{{5}}+{7}{e}^{{1}}-{3}{\left({1}\right)}-{0}-{7}{e}^{{0}}+{3}{\left({0}\right)}$$
$$\displaystyle=\frac{{1}}{{5}}+{7}{e}-{3}-{7}$$
$$\displaystyle=\frac{{1}}{{5}}+{7}{e}-{10}$$
$$\displaystyle={7}{e}-\frac{{{49}}}{{5}}$$
$$\displaystyle\therefore{\int_{{0}}^{{1}}}{\left({x}^{{4}}+{7}{e}^{{x}}-{3}\right)}{\left.{d}{x}\right.}={7}{e}-\frac{{49}}{{5}}$$