Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 20

Find the integral $\displaystyle \int^3_1 \left( 1 + 2x - 4 x^3 \right) dx$

$\begin{equation} \begin{aligned} \int \left( 1 + 2x - 4 x^3 \right) dx &= \int 1 dx + 2 \int x dx - 4 \int x^3 dx\\ \\ \int \left( 1 + 2x - 4 x^3 \right) dx &= x + 2 \left( \frac{x^{1+1}}{1+1} \right) - 4 \left( \frac{x^{3+1}}{3+1} \right) + C\\ \\ \int \left( 1 + 2x - 4 x^3 \right) dx &= x + \frac{\cancel{2}x^2}{\cancel{2}} - \frac{\cancel{4}x^4}{\cancel{4}} + C\\ \\ \int \left( 1 + 2x - 4 x^3 \right) dx &= x + x^2 - x^4 + C \end{aligned} \end{equation}$


$\begin{equation} \begin{aligned} \int^3_1 \left( 1 + 2x - 4 x^3 \right) dx &= 3 + (3)^2 - (3)^4 + C - \left[ 1 + (1)^2 - (1)^4 + C \right]\\ \\ \int^3_1 \left( 1 + 2x - 4 x^3 \right) dx &= 3 + 9 - 81 + C - 1 - 1 + 1 - C\\ \\ \int^3_1 \left( 1 + 2x - 4 x^3 \right) dx &= - 70 \end{aligned} \end{equation}$