Integrate ∫2−1(2x2+x)dx

Integrate ${\int }_{-1}^2(2x^2+x)dx$

Answer Details

${\int }_{-1}^2(2x^2+x)dx$
= $[\frac{2x^{2+1}}{3}+\frac{x^{1+1}}{2}{]}_{-1}^2$
= $[\frac{2x^3}{3}+\frac{x^2}{2}{]}_{-1}^2$
= $(\frac{2(2{)}^3}{3}+\frac{2^2}{2})-(\frac{2(-1{)}^3}{3}+\frac{(-1{)}^2}{2})$
= $(\frac{16}{3}+2)-(\frac{-2}{3}+\frac{1}{2})$
= $\frac{22}{3}-(-\frac{1}{6})$
= $\frac{22}{3}+\frac{1}{6}$
= $\frac{15}{2}$
= $7\frac{1}{2}$