Integrate

**PensFan10** · May 19th 2009, 06:10 AM

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

**Media_Man** · May 19th 2009, 07:07 AM

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

**Moo** · May 19th 2009, 10:54 AM

Hello;

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 $-\frac{x^{-\alpha}}{\alpha+1}$

**Media_Man** · May 19th 2009, 11:32 AM

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.

**PensFan10** · May 20th 2009, 04:50 AM

uhh that makes more sense. Thanks to both of you

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