Answer
$$ \frac{42}{5 }$$
Work Step by Step
Given
$$ \int_{-1}^{2} \sqrt{5 x+6} d x$$
Let
$$ u= 5x+6 \ \ \ \Rightarrow\ \ \ du = 5dx$$
At $ x=-1\to u= 1$ and $ x= 2 \to u= 16$
\begin{align*}
\int_{-1}^{2} \sqrt{5 x+6} d x &=\frac{1}{5} \int_{1}^{16} \sqrt{u} d u=\frac{1}{5} \int_{1}^{16} u^{1 / 2} d u \\ &=\left.\frac{1}{5}\left(\frac{2}{3} u^{3 / 2}\right)\right|_{1} ^{16} \\ &=\frac{2}{15}\left(16^{3 / 2}-1^{3 / 2}\right) \\ &=\frac{2}{15}(64-1)=\frac{2}{15}(63)\\
&= \frac{42}{5 }
\end{align*}
