Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 24

Find the integrals $\displaystyle \int^4_0 (2v + 5)(3v- 1) dv$

$
\begin{equation}
\begin{aligned}
\int (2v + 5)(3v- 1) dv &= \int \left( 6v^2 - 2v + 15v - 5 \right) dv\\
\\
\int (2v + 5)(3v- 1) dv &= \int \left( 6v^2 + 13v - 5 \right) dv \\
\\
\int (2v + 5)(3v- 1) dv &= 6 \int v^2 dv + 13 \int v d v - \int 5 dv\\
\\
\int (2v + 5)(3v- 1) dv &= 6 \left( \frac{v^{2+1}}{2+1} \right) + 13 \left( \frac{v^{1+1}}{1+1} \right) - 5 \left( \frac{v^{0+1}}{0+1} \right) + C\\
\\
\int (2v + 5)(3v- 1) dv &= 6 \left( \frac{v^3}{3} \right) + 13 \left( \frac{v^2}{2} \right) - 5v + C\\
\\
\int (2v + 5)(3v- 1) dv &= 2v^3 + \frac{13v^2}{2} - 5v + C\\
\\
\int^4_0 (2v + 5)(3v- 1) dv &= 2(4)^3 + \frac{13(4)^2}{2} - 5 (4) + C - \left[ 2(0)^3 + \frac{13(0)^2}{2} - 5 (0) + C \right]\\
\\
\int^4_0 (2v + 5)(3v- 1) dv &= 128 + 104 - 20 + C - 0 -0 + 0 - C\\
\\
\int^4_0 (2v + 5)(3v- 1) dv &= 212
\end{aligned}
\end{equation}
$