Single Variable Calculus, Chapter 5, 5.3, Section 5.3, Problem 30

Find the integral $\displaystyle \int^2_0 (y - 1)(2y + 1) dy$

Using 2nd Fundamental Theorem of Calculus

$\displaystyle \int^b_a f(x) dx = F(b) - F(a)$, where $F$ is any anti-derivative of $f$.

Let $\displaystyle f(y) = (y - 1)(2y + 1)$ or $f(y) = 2y^2 - y - 1$, then


$
\begin{equation}
\begin{aligned}

F(y) =& 2 \left( \frac{y^{2 + 1}}{2 + 1} \right) - \left( \frac{y^{1 + 1}}{1 + 1} \right) - 1 \left( \frac{y^{0 + 1}}{0 + 1} \right) + C
\\
\\
F(y) =& \frac{2y^3}{3} - \frac{y^2}{2} - y + C
\\
\\
F(y) =& \frac{2y^3}{3} - \frac{y^2}{2} - y + C

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

& \int^2_0 (y - 1)(2y + 1) dy = F(2) - F(0)
\\
\\
& \int^2_0 (y - 1)(2y + 1) dy = \frac{2(2)^3}{3} - \frac{(2)^2}{2} - 2 + C - \left[ \frac{2(0)^3}{3} - \frac{(0)^2}{2} - 0 + C \right]
\\
\\
& \int^2_0 (y - 1)(2y + 1) dy = \frac{16}{3} - 2 - 2 + C - 0 + 0 + 0 - C
\\
\\
& \int^2_0 (y - 1)(2y + 1) dy = \frac{16}{3} - 4
\\
\\
& \int^2_0 (y - 1)(2y + 1) dy = \frac{16 - 12}{3}
\\
\\
& \int^2_0 (y - 1)(2y + 1) dy = \frac{4}{3}

\end{aligned}
\end{equation}
$