Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 26

Find the integrals $\displaystyle \int^9_0 \sqrt{2t}dt$

$
\begin{equation}
\begin{aligned}
\int \sqrt{2t}dt &= \int \sqrt{2} \sqrt{t} dt\\
\\
\int \sqrt{2t}dt &= \sqrt{2} \int \sqrt{t} dt \\
\\
\int \sqrt{2t}dt &= \sqrt{2} \int (t)^{\frac{1}{2}} dt \\
\\
\int \sqrt{2t}dt &= \sqrt{2} \left( \frac{t^{\frac{1}{2}+1 }}{\frac{1}{2}+1} \right) + C\\
\\
\int \sqrt{2t}dt &= \frac{\sqrt{2}(t)^{\frac{3}{2}}}{\frac{3}{2}} + C\\
\\
\int \sqrt{2t}dt &= \frac{2\sqrt{2t^3}}{3} + C\\
\\
\int^9_0 \sqrt{2t}dt &= \frac{2\sqrt{2(9)^3}}{3} + C - \left( \frac{2\sqrt{2(0)^3}}{3} + C \right)\\
\\
\int^9_0 \sqrt{2t}dt &= \frac{2(27\sqrt{2})}{3} + C - 0 - C\\
\\
\int^9_0 \sqrt{2t}dt &= 18\sqrt{2}
\end{aligned}
\end{equation}
$