# Integrate

• May 19th 2009, 07:10 AM
PensFan10
Integrate
The bottom limit is beta and the top limit is y the function is 1/x^(alpha +1) dx
• May 19th 2009, 08:07 AM
Media_Man
$\int_\beta^y \frac{1}{x^{\alpha+1}} dx$ = $\int_\beta^y x^{-\alpha-1} dx$ = $-\frac{x^{-\alpha}}{\alpha+1} |_{x=\beta}^y$ = $-\frac{y^{-\alpha}}{\alpha+1}+\frac{\beta^{-\alpha}}{\alpha+1}$= $\frac{\beta^{-\alpha}-y^{-\alpha}}{\alpha+1}$= $\frac{y^\alpha-\beta^\alpha}{(\alpha+1)(y\beta)^\alpha}$
• May 19th 2009, 11:54 AM
Moo
Hello;
Quote:

Originally Posted by Media_Man
$\int_\beta^y \frac{1}{x^{\alpha+1}} dx$ = $\int_\beta^y x^{-\alpha-1} dx$ = $-\frac{x^{-\alpha}}{\alpha+1} |_{x=\beta}^y$ = $-\frac{y^{-\alpha}}{\alpha+1}+\frac{\beta^{-\alpha}}{\alpha+1}$= $\frac{\beta^{-\alpha}-y^{-\alpha}}{\alpha+1}$= $\frac{y^\alpha-\beta^\alpha}{(\alpha+1)(y\beta)^\alpha}$

Hmmm except that an antiderivative of $x^{-\alpha-1}$ is $-\frac{x^{-\alpha}}{\alpha}$, and not $-\frac{x^{-\alpha}}{\alpha+1}$ (Worried)
• May 19th 2009, 12:32 PM
Media_Man
Right you are
Quote:

Originally Posted by Media_Man
$\int_\beta^y \frac{1}{x^{\alpha+1}} dx$ = $\int_\beta^y x^{-\alpha-1} dx$ = $-\frac{x^{-\alpha}}{\alpha} |_{x=\beta}^y$ = $-\frac{y^{-\alpha}}{\alpha}+\frac{\beta^{-\alpha}}{\alpha}$= $\frac{\beta^{-\alpha}-y^{-\alpha}}{\alpha}$= $\frac{y^\alpha-\beta^\alpha}{\alpha(y\beta)^\alpha}$

*Yes, my bad. Quick fix, though.
• May 20th 2009, 05:50 AM
PensFan10
uhh that makes more sense. Thanks to both of you