What am I doing wrong? (indefinite integrals)

**birdz** · April 19th 2009, 04:13 PM

I've been trying to finish my hw on web assign, but can't get this problem right. I've tried reading my book, but so far as I can tell I'm doing it the way it says to. I also checked by differentiation, and I thought it was right.

(this is a screencap of the problem, and my solution). Thanks for your help

**Calculus26** · April 19th 2009, 04:18 PM

u = x^3 +28 du = 3x^2dx

which is where the 1/3 comes from

integral 0f U^1/2 is 2/3U^3/2 so your coefficient is 2/9

**skeeter** · April 19th 2009, 04:20 PM

Originally Posted by birdz

I've been trying to finish my hw on web assign, but can't get this problem right. I've tried reading my book, but so far as I can tell I'm doing it the way it says to. I also checked by differentiation, and I thought it was right.

(this is a screencap of the problem, and my solution). Thanks for your help

$\frac{1}{3} \int 3x^2 \sqrt{x^3 + 28} \, dx$

$\frac{1}{3} \cdot \frac{2}{3}\left(x^3 + 28\right)^{\frac{3}{2}} + C$

**mollymcf2009** · April 19th 2009, 04:28 PM

Originally Posted by birdz

I've been trying to finish my hw on web assign, but can't get this problem right. I've tried reading my book, but so far as I can tell I'm doing it the way it says to. I also checked by differentiation, and I thought it was right.

(this is a screencap of the problem, and my solution). Thanks for your help

$u = x^3 + 28$

$du = 3x^2 dx$

$\frac{1}{3} du = x^2 dx$

$\frac{1}{3} \int \sqrt{u} du$

$= \frac{1}{3} \int u^{\frac{1}{2}} du$

Then integrate:

$\frac{1}{3} \left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right] + C$

$= \frac{1}{3}\left[\frac{2}{3} u^{\frac{3}{2}}\right] + C$

Substitute back in for u, just like you did.

You just left out the $\frac{2}{3}$ from your integration. Try it with that, otherwise your answer is in the right form.

**birdz** · April 19th 2009, 04:29 PM

Thank you so much!

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