Answer
The given improper integral
$$
\int_{2}^{6} \frac{y d y}{\sqrt{y-2}} = \frac{40}{3}
$$
Work Step by Step
$$
\int_{2}^{6} \frac{y d y}{\sqrt{y-2}}
$$
First, we evaluate the Indefinite Integral
$$
\int \frac{y d y}{\sqrt{y-2}}
$$
Let $ u= \sqrt{y-2} $. Then $ u^{2}= y-2 $ and $ 2u du = dy $, so
$$
\begin{aligned}
\int \frac{y d y}{\sqrt{y-2}} &=\int \frac{\left(u^{2}+2\right) 2 u d u}{u}\\
&=2 \int\left(u^{2}+2\right) d u\\
&=2\left[\frac{1}{3} u^{3}+2 u\right]+C \\
&= \left[\frac{2}{3}(y-2)^{3 / 2}+4 \sqrt{y-2}\right] +C \quad\quad\quad (1)
\end{aligned}
$$
Now, we evaluate the given definite integral:
$$
\int_{2}^{6} \frac{y d y}{\sqrt{y-2}}
$$
Observe that the given integral is improper because $ f(y)= \frac{y }{\sqrt{y-2}}$ has the vertical asymptote $y =2 $. Since the infinite discontinuity occurs at the left endpoint of $[2, 6] $, we must use part (b) of Definition 3 :
Thus, from formula (1) we get
$$
\begin{aligned} \int_{2}^{6} \frac{y d y}{\sqrt{y-2}} &=\lim _{t \rightarrow 2^{+}} \int_{t}^{6} \frac{y d y}{\sqrt{y-2}}\\
&=\lim _{t \rightarrow 2^{+}}\left[\frac{2}{3}(y-2)^{3 / 2}+4 \sqrt{y-2}\right]_{t}^{6} \\ &=\lim _{t \rightarrow 2^{+}}\left[\frac{16}{3}+8-\frac{2}{3}(t-2)^{3 / 2}-4 \sqrt{t-2}\right]=\frac{40}{3} \end{aligned}
$$
Thus the given improper integral
$$
\int_{2}^{6} \frac{y d y}{\sqrt{y-2}} = \frac{40}{3}
$$
