# Evaluate the integrals. int_-2^2(3x^4-2x+1)dx

Evaluate the integrals.
${\int }_{-2}^{2}\left(3{x}^{4}-2x+1\right)dx$
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Step 1:Given that
Evaluate integrals.
${\int }_{-2}^{2}\left(3{x}^{4}-2x+1\right)dx$
Step 2:Solving the integral
We have,
${\int }_{-2}^{2}\left(3{x}^{4}-2x+1\right)dx={\int }_{2}^{23}{x}^{4}dx-2{\int }_{-2}^{2}xdx+{\int }_{-2}^{21}dx$
$=3{\left[\frac{{x}^{5}}{5}\right]}_{-2}^{2}-2{\left[\frac{{x}^{2}}{2}\right]}_{-2}^{2}+{\left[x\right]}_{-2}^{2}$
$=\frac{3}{5}\left[{2}^{5}-{\left(-2\right)}^{5}\right]-\left[{2}^{2}-{\left(-2\right)}^{2}\right]+\left[2-\left(-2\right)\right]$
$=\frac{3}{5}\left(32-\left(-32\right)\right)-\left(4-4\right)+\left(2+2\right)$
$\frac{3}{5}\left(64\right)-0+4$
$=\frac{192}{5}+4$
$=\frac{192+20}{5}$
$=\frac{212}{5}=42.4$