# Thread: stuck on an integral

1. ## stuck on an integral

49
S dx/ (27+2x)^(2/3)
0

I'm pretty confused on what to do, help me figure it out please.

2. u = 27+2x --> du = 2 dx
change the value of upper/lower limits
and it will be easy O_o

3. I understand changing the limits but I can't figure out how to integrate this properly, I keep getting an incorrect answer

4. $\int_0^{49}\frac{1}{(27+2x)^{\frac{2}{3}}}dx$

So u = 27 + 2x

du = 2dx
dx = du/2

change the limits, 27 to 125

$\int_{27}^{125}\frac{1}{2(u)^{\frac{2}{3}}}du$

$\int_{27}^{125}\frac{1}{2}*u^{\frac{-2}{3}}du$

power rule:

$\frac{3}{2}*u^{\frac{1}{3}}$ from 27 to 125, can you take it from there?

5. hmm that is exactly what I've been doing but when I input my answer through my online hw on webassign: (3/2)(277^(1/3)-3) it tells me I'm wrong. :/ anything wrong with that answer?

6. $\frac{3}{2}*u^{\frac{1}{3}}
$

$\frac{3}{2}*(125^{\frac{1}{3}})-\frac{3}{2}*(27^{\frac{1}{3}})$

$\frac{3}{2}*5-\frac{3}{2}*3$

I don't see where you get this 277 from.

7. do you not replace u by (27+2x)?

8. well you changed your limits to accomodate for u right?

x goes from 0 to 49.

u= 27 + 2x

x = 0 => u=27 + 2*0 = 27

x = 49 => u=27+2*49=27+98=125

9. we covered the limits several replies ago, but 3 was correct thanks selb, I'm not sure where I got the idea i needed to throw 27+2x back in there

10. I'm still not sure where you get this 277 from but the point of the last question was to check that you followed why I simply substituted 125 and 27 in for u instead of replacing it with the 27+2x, because i'm lazy and don't feel like substituting back in if I already know what it's going to be.

But here is the solution with substituting the 27+2x back in:

$\frac{3}{2}*u^{\frac{1}{3}}
$
u goes from 27 to 125

$\frac{3}{2}*(27+2x)^{\frac{1}{3}}$ x goes from 0 to 49

$=\frac{3}{2}*(27+2(49))^{\frac{1}{3}}-\frac{3}{2}*(27+2(0))^{\frac{1}{3}}$

$=\frac{3}{2}*(27+98)^{\frac{1}{3}}-\frac{3}{2}*(27)^{\frac{1}{3}}$

$=\frac{3}{2}*(125)^{\frac{1}{3}}-\frac{3}{2}*(27)^{\frac{1}{3}}$

$=\frac{3}{2}*5-\frac{3}{2}*3$

does that help?