# Hard (or maybe it's not?) u-substitution definite integral problem

• Nov 21st 2010, 08:21 AM
chickeneaterguy
Hard (or maybe it's not?) u-substitution definite integral problem
$\int^1_{-1} t^3 (1+ t^4)^3 dx$

How do I go about solving this one? There are t's but it has dx

I made $u = 1 + t^4$
but $\frac {du}{dt}$ isn't equal to $4t^3$, right?

What do I do?

Thanks, chickeneaterguy
• Nov 21st 2010, 08:25 AM
Also sprach Zarathustra
You can solve it without any kind of substitution, take a look at the integrand...
• Nov 21st 2010, 08:28 AM
chickeneaterguy
it's on my practice exam and I'm required to solve it with u-sub :(
• Nov 21st 2010, 08:30 AM
Unknown008
I think that the question made a typo and put dx istead of dt.

And du/dt is indeed 4t^3

Don't forget to change the limits when you make the substitution.
• Nov 21st 2010, 08:30 AM
Hardwork
Quote:

Originally Posted by Also sprach Zarathustra
You can solve it without any kind of substitution, take a look at the integrand...

I know what you are thinking, but the integral is with respect to x, not t. (Thinking)
• Nov 21st 2010, 08:34 AM
chickeneaterguy
Quote:

Originally Posted by Unknown008
I think that the question made a typo and put dx istead of dt.

And du/dt is indeed 4t^3

Don't forget to change the limits when you make the substitution.

That's what I thought too as he hasn't taught us how to do this. He has taught us $\frac {dx}{dt}$ and stuff like that but if I do it that way, it doesn't give me a specific value. I look at the graph and it's apparent that the area should be zero, but yeah...must be a typo.

Quote:

Originally Posted by Hardwork
I know what you are thinking, but the integral is with respect to x, not t. (Thinking)

This is just Calc 1, have you seen problem/s like this before?
• Nov 21st 2010, 08:36 AM
Also sprach Zarathustra
It is dt and not dx!

If it was dx the answer is 2t^3(1+t^4)^3
• Nov 21st 2010, 08:38 AM
Unknown008
Quote:

Originally Posted by Also sprach Zarathustra
It is dt and not dx!

If it was dx the answer is 2t^3(1+t^4)^3

I'm curious about how you got that... I was thinking that if it were indeed dx, then you'd have this expression as a constant... or not?
• Nov 21st 2010, 08:41 AM
Also sprach Zarathustra
Quote:

Originally Posted by chickeneaterguy
it's on my practice exam and I'm required to solve it with u-sub :(

I don't see the reason why...
• Nov 21st 2010, 08:42 AM
Also sprach Zarathustra
Quote:

Originally Posted by chickeneaterguy
it's on my practice exam and I'm required to solve it with u-sub :(

I don't see the reason why...
• Nov 21st 2010, 08:48 AM
Hardwork
Quote:

Originally Posted by Also sprach Zarathustra
I don't see the reason why...

Hey, I'm curious too. How did you get that answer when it's integrated with respect to x?
• Nov 21st 2010, 08:48 AM
chickeneaterguy
I just did it with $\frac {du}{dt}$, thanks for helping though. :)
• Nov 21st 2010, 11:01 AM
minusb
Quote:

Originally Posted by chickeneaterguy
I just did it with $\frac {du}{dt}$, thanks for helping though. :)

How did you go about it in the end?
• Nov 21st 2010, 11:09 AM
chickeneaterguy
Quote:

Originally Posted by minusb
How did you go about it in the end?

I just assumed the dx was a typo and was supposed to be a dt.

I picked $u=1+t^4$ as my u and $\frac {du}{dt} = 4t^3$ therefore $du= 4t^3 dt$. I then plugged my 1 and -1 from the integral into $u = 1+t^4$ to get $\int^2_2 \frac 14u^3 du$ which when integrated is $\frac 1{16}u^4$ at 2 and 2, so $\frac 1{16}2^4 - \frac 1{16}2^4 = 0$