Integrate

• May 19th 2009, 06:10 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, 07:07 AM
Media_Man
$\displaystyle \int_\beta^y \frac{1}{x^{\alpha+1}} dx$ =$\displaystyle \int_\beta^y x^{-\alpha-1} dx$ =$\displaystyle -\frac{x^{-\alpha}}{\alpha+1} |_{x=\beta}^y$ =$\displaystyle -\frac{y^{-\alpha}}{\alpha+1}+\frac{\beta^{-\alpha}}{\alpha+1}$=$\displaystyle \frac{\beta^{-\alpha}-y^{-\alpha}}{\alpha+1}$=$\displaystyle \frac{y^\alpha-\beta^\alpha}{(\alpha+1)(y\beta)^\alpha}$
• May 19th 2009, 10:54 AM
Moo
Hello;
Quote:

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

Hmmm except that an antiderivative of $\displaystyle x^{-\alpha-1}$ is $\displaystyle -\frac{x^{-\alpha}}{\alpha}$, and not $\displaystyle -\frac{x^{-\alpha}}{\alpha+1}$ (Worried)
• May 19th 2009, 11:32 AM
Media_Man
Right you are
Quote:

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

*Yes, my bad. Quick fix, though.
• May 20th 2009, 04:50 AM
PensFan10
uhh that makes more sense. Thanks to both of you