1. ## Integrate

The bottom limit is beta and the top limit is y the function is 1/x^(alpha +1) dx

2. $\displaystyle \int_\beta^y \frac{1}{x^{\alpha+1}} dx$ =$\displaystyle \int_\beta^y x^{-\alpha-1} dx$ =$\displaystyle -\frac{x^{-\alpha}}{\alpha+1} |_{x=\beta}^y$ =$\displaystyle -\frac{y^{-\alpha}}{\alpha+1}+\frac{\beta^{-\alpha}}{\alpha+1}$=$\displaystyle \frac{\beta^{-\alpha}-y^{-\alpha}}{\alpha+1}$=$\displaystyle \frac{y^\alpha-\beta^\alpha}{(\alpha+1)(y\beta)^\alpha}$

3. Hello;
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 $\displaystyle -\frac{x^{-\alpha}}{\alpha+1}$

4. ## Right you are

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.

5. uhh that makes more sense. Thanks to both of you