evaluate: integral from 3 to infinity: 1/(x-2)^(3/2) dx

Question

anonymous · Answer

$$\int\limits_{3}^{\infty?} (1)\div ((x-2)^{(3/2)}) dx$$

anonymous · Answer

$$\lim_{n \rightarrow \inf} \int\limits_{3}^{n}(1/((n-2)^(3/2))dn$$ 

= $$\lim_{n \rightarrow \inf} [-2/(\sqrt{n-2}) - (-2/\sqrt{3-2} )$$

=0+2=2

anonymous · Answer

I skipped all of the integral steps and went straight to evaluating them... let me know if you have questions on the in between steps.

anonymous · Answer

ah yes! thats the part i got stuckk! thank u :)

anonymous · Answer

so you got it?

anonymous · Answer

where is the -2 in the numerator coming from?

anonymous · Answer

$$\int_3^\infty (x-2)^{-3/2} dx=\frac{(x-2)^{-1/2}}{-1/2}+C$$
$$(x-2)^{-1/2}=\frac{1}{\sqrt{x-2}}$$, which approaches 0 as x approaches infinity.
Hence the integral is $$0-\frac{(3-2)^{-1/2}}{-1/2}=2$$

anonymous · Answer

I used u substitution for the integral and integral of u^(-3/2) = -2/sqrt(u)