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

  1. #1
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    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
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    You can solve it without any kind of substitution, take a look at the integrand...
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  3. #3
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    it's on my practice exam and I'm required to solve it with u-sub
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  4. #4
    MHF Contributor Unknown008's Avatar
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    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.
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  5. #5
    Junior Member Hardwork's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    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.
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  6. #6
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    Quote Originally Posted by Unknown008 View Post
    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 View Post
    I know what you are thinking, but the integral is with respect to x, not t.
    This is just Calc 1, have you seen problem/s like this before?
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    It is dt and not dx!

    If it was dx the answer is 2t^3(1+t^4)^3
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    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?
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  9. #9
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by chickeneaterguy View Post
    it's on my practice exam and I'm required to solve it with u-sub

    I don't see the reason why...
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  10. #10
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by chickeneaterguy View Post
    it's on my practice exam and I'm required to solve it with u-sub

    I don't see the reason why...
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  11. #11
    Junior Member Hardwork's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    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?
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  12. #12
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    I just did it with  \frac {du}{dt} , thanks for helping though.
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  13. #13
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    Quote Originally Posted by chickeneaterguy View Post
    I just did it with  \frac {du}{dt} , thanks for helping though.
    How did you go about it in the end?
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  14. #14
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    Quote Originally Posted by minusb View Post
    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
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