Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 22

Find the integrals $\displaystyle \int^0_{-2} \left( u^5 - u^3 + u^2 \right) du$

$\begin{equation} \begin{aligned} \int \left( u^5 - u^3 + u^2 \right) du &= \int u^5 du - \int u^3 du + \int u^2 du\\ \\ \int \left( u^5 - u^3 + u^2 \right) du &= \frac{u^{5+1}}{5+1} - \frac{u^{3+1}}{3+1} + \frac{u^{2+1}}{2+1} + C\\ \\ \int \left( u^5 - u^3 + u^2 \right) du &= \frac{u^6}{6} - \frac{u^4}{4} + \frac{u^3}{3} + C\\ \\ \int^0_{-2} \left( u^5 - u^3 + u^2 \right) du &= \frac{(0)^6}{6} - \frac{(0)^4}{4} + \frac{(0)^3}{3} + C - \left[ \frac{(-2)^6}{6} - \frac{(-2)^4}{4} + \frac{(-2)^3}{3} + C\right]\\ \\ \int^0_{-2} \left( u^5 - u^3 + u^2 \right) du &= C - \frac{64}{6} + \frac{16}{4} + \frac{8}{3} - C\\ \\ \int^0_{-2} \left( u^5 - u^3 + u^2 \right) du &= -4 \end{aligned} \end{equation}$