calculate integral $\;2\int_{-2}^{0} \sqrt{8x+16}dx$ I want to calculate the integral  $$2\int_{-2}^{0} \sqrt{8x+16}dx$$ The answer is $\;\dfrac {32}{6}\;$ but I don't know how to get it.  
 A: $$I = F(x) = \int_{-2}^0 \sqrt{(8x + 16)}\,dx$$
We use substitution: 


*

*Let $u = 8x + 16,\;\;du = 8\,dx \implies dx = \dfrac 18 du$


Change limits of integration: 


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*When $x = -2, u = 0$, when $x = 0, u = 16$. 


Substituting equivalent expressions and changing the limits of integration then gives us:
$$\int_{-2}^0 \sqrt{(8x + 16)}\,dx=\int_{0}^{16} \sqrt{u}\,\left(\frac 18 du\right) = \dfrac 18 \int_0^{16} u^{1/2}\,du$$
Now we use the power rule to integrate: $\quad \int u^a\,du = \dfrac{u^{a + 1}}{a+1} + C,\quad\text{for all}\;a\neq -1$
We integrate with respect to $u$ and evaluate the result $I = F(u)$: $F(16) - F(0)$.
$$ \dfrac 18 \int_0^{16} u^{1/2}\,du = \frac{1}{8}\left(\frac{2}{3}u^{3/2}\right)=\frac{1}{12}\left(u^{3/2}\right)\Bigg|^{16}_{0} = \frac{1}{12}\Bigl[(64) - (0)\Bigr] = \frac{32}{6}$$

If your integral was, as you write it, given as $2I = 2F(x) = 2F(u)$, then our result will be $$2\cdot \frac {32}{6} = \frac{32}{3}$$
A: Make the substitution, $y=8x+16$. This gives us $dx = dy/8$. Hence, the integral becomes
$$I=\int_0^{16}\sqrt{y} dy/8$$
I trust you can finish it from here, by recall the formula $\displaystyle\int y^n dy = \dfrac{y^{n+1}}{n+1} + \text{constant}$
A: Use a simple $u$ substitution. Let $u = 8x + 16$. Then $du = 8dx$, and we have:
$$\int_{-2}^{0} \sqrt{8x+16}dx = \frac{1}{8}\int_{0}^{16}u^{1/2}du = \frac{1}{8}[\frac{2}{3}u^{3/2}|^{16}_{0}] = \frac{1}{12}[(64) - (0)] = \frac{32}{6}$$
