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

Step 1: Given that
Evaluate integrals.
${\displaystyle {\int }_{-2}^2(3x^4-2x+1)dx}$
Step 2: Solving the integral
We have,
$K{\int }_{-2}^2(3x^4-2x+1)dx={\int }_2^{2}3x^{4}dx-2{\int }_{-2}^{2}xdx+{\int }_{-2}^{2}1dx$${\displaystyle =3{\left[\frac{x^5}{5}\right]}_{-2}^2-2{\left[\frac{x^2}{2}\right]}_{-2}^2+{\left[x\right]}_{-2}^2}$
${\displaystyle =\frac{3}{5}[2^5-{(-2)}^5]-[2^2-{(-2)}^2]+[2-(-2)]}$
${\displaystyle =\frac{3}{5}(32-(-32))-(4-4)+(2+2)}$
${\displaystyle =\frac{3}{5}\left(64\right)-0+4}$
${\displaystyle =\frac{192}{5}+4}$
${\displaystyle =\frac{192+20}{5}}$
${\displaystyle =\frac{212}{5}}$
=42.4