# Math Help - [SOLVED] Help needed with a definite integral

1. ## [SOLVED] Help needed with a definite integral

$
\int_0^1 x^2e^{x^{1/5}}~dx
$

I assume that this is solved using integration by parts, but what should I choose for u and dv?

2. I'm not sure on this but I think you can do integration by parts if you make a substituion.

$w = x^{\frac{1}{5}}$

3. Originally Posted by 11rdc11
I'm not sure on this but I think you can do integration by parts if you make a substituion.

$w = x^{\frac{1}{5}}$
yeah, but it would be crazy! you would have to do it like 14 times. the resulting integral is:

$5 \int u^{14}e^u~du$

4. How would that work? Would I just solve the problem like normal, but totally ignore the x^1/5 part until the very end, and simply swap w and x^1/5?

5. Originally Posted by Jhevon
yeah, but it would be crazy! you would have to do it like 14 times. the resulting integral is:

$5 \int u^{14}e^u~du$
Lol yea I just noticed. Umm can't we use a reduction formula after?

6. Originally Posted by redman223
How would that work? Would I just solve the problem like normal, but totally ignore the x^1/5 part until the very end, and simply swap w and x^1/5?
Yep you would just solve the problem like normal and you don't have to worry bout swapping $x^{\frac{1}{5}}$ for w because all you have to do is change your lower and upper bound to correlate with w.

For example

$0^{\frac{1}{5}} = 0$

and upper bound

$1^{\frac{1}{5}} = 1$

$5 \int_0^1 w^{14}e^w~dw$

7. So what would be u in this case? Would that be the same thing as the w mentioned before?

8. Oops made corrections. I used variable w so I wouldn't get mixed up when doing integration by parts, since integration by parts normally uses u and v.

9. Just put this integral in the TI89 and the solution is very long

10. Originally Posted by 11rdc11
Just put this integral in the TI89 and the solution is very long
yup. when i saw the integral, i thought it would be messy so i put it into the integrator, the solution there was messy as well.

i hope there is a mistake in this problem

maybe there is a shortcut. we can wait on Kriz or one of the other integral people to see

11. $\int_0^1 x^2 e^{x^{1/5}}~dx ~=~ 5\int_0^1 u^{14} e^u~du$

$\int u^{14}e^u~du = u^{14} e^u - 14 u^{13}e^u + 14.13 u^{12}e^u ..... - 14! u^1 e^u+14! e^u~+$ $C = \sum_{i=0}^{14}u^i e^u (-1)^i \frac{14!}{i!}~+C$

Evaluating this from 0 to 1 and multiplying by 5 gives us $5\left [ \left ( \sum_{i=0}^{14} e (-1)^i \frac{14!}{i!} \right )-14! \right ] \approx 0.8526185$