# Thread: Improper Integrals

1. ## Improper Integrals

I'm having trouble with the following problem:

Show that $\displaystyle \int_{1}^{\infty} \frac {du}{u^\frac{3}{2}}$

2. Originally Posted by larson
I'm having trouble with the following problem:

Show that $\displaystyle \int_{1}^{\infty} \frac {du}{u^\frac{3}{2}}$
Show that this is what?

$\displaystyle \int_{1}^{\infty} \frac {du}{u^\frac{3}{2}}$

$\displaystyle = \lim_{b \to \infty} \int_{1}^b \frac {du}{u^\frac{3}{2}}$

$\displaystyle = \lim_{b \to \infty}-2 \left ( \frac{1}{b^{1/2}} - \frac{1}{1^{1/2}} \right )$

$\displaystyle = 2$

-Dan

3. Originally Posted by topsquark
Show that this is what?

$\displaystyle \int_{1}^{\infty} \frac {du}{u^\frac{3}{2}}$

$\displaystyle = \lim_{b \to \infty} \int_{1}^b \frac {du}{u^\frac{3}{2}}$

$\displaystyle = \lim_{b \to \infty}-2 \left ( \frac{1}{b^{1/2}} - \frac{1}{1^{1/2}} \right )$

$\displaystyle = 2$

-Dan
How do you know to take the 2 out? Thats the only thing thats confusing me, and by the way the answer is correct, although I'm sure you knew .

4. Hello
$\displaystyle \int x^{\alpha} \,\mathrm{d}x= \frac{t^{\alpha+1}}{\alpha+1}$ with $\displaystyle \alpha \neq -1$
Applying it with $\displaystyle \alpha=-\frac{3}{2}$ gives us :$\displaystyle \int \frac{1}{x^{\frac{3}{2}}} \,\mathrm{d}x= \int x^{-\frac{3}{2}} \,\mathrm{d}x = \frac{t^{-\frac{3}{2}+1}}{\frac{-3}{2}+1}=\frac{t^{-\frac{1}{2}}}{-\frac{1}{2}}=-\frac{2}{\sqrt{t}}$