∫−11(x+1)2dx≡∫−11(x2+2x+1)dx\int\limits_{-1}^{1} (x + 1)^{2} d x \equiv \int\limits_{-1}^{1} (x^{2} + 2x + 1) d x−1∫1(x+1)2dx≡−1∫1(x2+2x+1)dx
= x33+x2+x∣−11\; \frac{x^{3}}{3} + x^{2} + x \big|_{-1}^{1}3x3+x2+x−11
= ( 133+12+1)−( (−1)33+(−1)2+(−1))= 73+ 13= 83\left(\; \frac{1^{3}}{3} + 1^{2} + 1\right) - \left(\; \frac{(-1)^{3}}{3} + (-1)^{2} + (-1)\right) = \; \frac{7}{3} + \; \frac{1}{3} = \; \frac{8}{3}(313+12+1)−(3(−1)3+(−1)2+(−1))=37+31=38
