# How do you evaluate the integral int dx/sqrt(3-x) from 1 to 3 if it converges?

Jul 25, 2016

$2 \sqrt{2}$

#### Explanation:

We have:

${\int}_{1}^{3} \frac{\mathrm{dx}}{\sqrt{3 - x}}$

We should try substitution, namely by letting $u = 3 - x$. This implies that $\mathrm{du} = - \mathrm{dx}$. The bounds will also change, with $1$ becoming $3 - 1 = 2$ and $3$ becoming $3 - 3 = 0$.

${\int}_{1}^{3} \frac{\mathrm{dx}}{\sqrt{3 - x}} = - {\int}_{2}^{0} \frac{\mathrm{du}}{\sqrt{u}}$

We can use fractional and negative exponents in the integrand, as well as switching the direction of the bounds since we have the negative sign out front.

$- {\int}_{2}^{0} \frac{\mathrm{du}}{\sqrt{u}} = {\int}_{0}^{2} {u}^{- \frac{1}{2}} \mathrm{du}$

Integrate using the rule: ${\int}_{a}^{b} {u}^{n} \mathrm{du} = {\left[{u}^{n + 1} / \left(n + 1\right)\right]}_{a}^{b}$

${\int}_{0}^{2} {u}^{- \frac{1}{2}} \mathrm{du} = {\left[{u}^{- \frac{1}{2} + 1} / \left(- \frac{1}{2} + 1\right)\right]}_{0}^{2} = {\left[{u}^{\frac{1}{2}} / \left(\frac{1}{2}\right)\right]}_{0}^{2} = {\left[2 \sqrt{u}\right]}_{0}^{2} = 2 \sqrt{2}$

Jul 25, 2016

$2 \sqrt{2}$

#### Explanation:

${\int}_{1}^{3} \frac{1}{\sqrt{3 - x}} \setminus \mathrm{dx}$

we have a pattern here from the power and chain rules

$\frac{d}{\mathrm{dx}} \left(\sqrt{3 - x}\right)$
$= \frac{1}{2} \cdot \frac{1}{\sqrt{3 - x}} \cdot \left(- 1\right) = - \frac{\frac{1}{2}}{\sqrt{3 - x}}$

so our integrand is $- 2 \frac{d}{\mathrm{dx}} \left(\sqrt{3 - x}\right)$

and so we are integrating

${\int}_{1}^{3} \setminus - 2 \frac{d}{\mathrm{dx}} \left(\sqrt{3 - x}\right) \mathrm{dx}$

$= - 2 {\left[\sqrt{3 - x}\right]}_{1}^{3} = 2 {\left[\sqrt{3 - x}\right]}_{3}^{1}$

$= 2 \left(\sqrt{2} - 0\right) = 2 \sqrt{2}$