Determining if the integral $\int_2^5 \frac1{\sqrt{x-2}}dx$ converges or diverges

Question

Does the integral $$\int_2^5 \frac{1}{\sqrt{x-2}}dx$$ converge or diverge? Do i set $u=x-2$ to the denominator then find new bounds and find the limit?

Answer 1

I'll work through the problem as you suggested (your suggestion was mostly right :) good job!) although instead of letting $u$ equal $\sqrt{x-2}$, just let it equal $u$.

Let $u = x-2$. Then $du = dx$ and when $x = 2$, $u = 0$ and when $x = 5$, $u = 3$, so we have

$$\int \limits_{2}^{5} \frac{1}{\sqrt{x-2}}\,dx = \int \limits_{0}^{3} \frac{1}{\sqrt{u}}\,du = \int \limits_{0}^{3} u^{-\frac{1}{2}}\,dx = \frac{u^{\frac{1}{2}}}{(\frac{1}{2})}|_{0}^{3} = 2u^{\frac{1}{2}}|_{0}^{3} = 2 \cdot \sqrt{3}. $$

So, $\int \limits_{2}^{5} \frac{1}{\sqrt{x-2}}\,dx = 2\sqrt{3}$.

Answer 2

Directly:

$$\left.\int_2^5\frac{dx}{\sqrt{x-2}}=2\sqrt{x-2}\right|_2^5=2\sqrt3-\lim_{x\to2^+}2\sqrt{x-2}=2\sqrt3$$

and thus the integral converges and that's its value.