∫−12(2x2+x) dx\int\limits_{-1}^{2} (2x^2 + x) \, dx−1∫2(2x2+x)dx
= [2x2+13+x1+12]−12\left[ \frac{2x^{2+1}}{3} + \frac{x^{1+1}}{2} \right]_{-1}^{2}[32x2+1+2x1+1]−12
= [2x33+x22]−12\left[ \frac{2x^{3}}{3} + \frac{x^{2}}{2} \right]_{-1}^{2}[32x3+2x2]−12
= (2(2)33+222)−(2(−1)33+(−1)22)\left( \frac{2(2)^{3}}{3} + \frac{2^{2}}{2} \right) - \left( \frac{2(-1)^{3}}{3} + \frac{(-1)^{2}}{2} \right)(32(2)3+222)−(32(−1)3+2(−1)2)
= (163+2)−(−23+12)\left( \frac{16}{3} + 2 \right) - \left( \frac{-2}{3} + \frac{1}{2} \right)(316+2)−(3−2+21)
= 223−(−16)\frac{22}{3} - \left( -\frac{1}{6} \right)322−(−61)
= 223+16\frac{22}{3} + \frac{1}{6}322+61
= 152\frac{15}{2}215
= 7127\frac{1}{2}721