MHB Is this integral substitution approach correct for evaluating the integral I?

[karush]

$\large{S6.7.r.44}$
$$\displaystyle
I=\int_{2}^{6}\frac{y}{\sqrt{y-2}} \,dy = \frac{40}{3}$$
$$
\begin{align}
u&=y-2 &y&=u+2 \\
du&=dy
\end{align}$$
then
$$\displaystyle
I=\int_{0}^{4}\frac{u+2}{\sqrt{u}} \, du
=\int_{0}^{4}{u}^{1/2} \, du + 2\int_{0}^{4} {u}^{-1/2} \, du$$
Just seeing if going in right direction...

 

[Greg]

Looks good.

 

[karush]

$\large{S6.7.r.44}$
$$\displaystyle
I=\int_{2}^{6}\frac{y}{\sqrt{y-2}} \,dy = \frac{40}{3}$$
$$
\begin{align}
u&=y-2 &y&=u+2 \\
du&=dy
\end{align}$$
then
$$\displaystyle
I=\int_{0}^{4}\frac{u+2}{\sqrt{u}} \, du
=\int_{0}^{4}{u}^{1/2} \, du
+ 2\int_{0}^{4} {u}^{-1/2} \, du
=\frac{ \sqrt{2}u(u+4)}{4 } \\
\text{back subst and calc gives} \\
I=\frac{40}{3}$$