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Math Help - Integral Help

  1. #1
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    Exclamation Integral Help

    \int5x\sqrt[3]{1-x^2}dx

    I am required to use U subsitiution.

    u=  1-x^2
    du= -2x dx

    So the problem became....

    \frac{-2}{5} \int u^{1/3} du

    and my final answer was...

    \frac{-3}{10} u^{4/3} + C


    This seems correct to me, however the answer my book gives is...
    \frac{-15}{8} u^{4/3} + C

    Of course I will replace the u in the final answer, but I cannot figure out how the book did this problem. Can someone please tell me what I did wrong.

    Thanks so much!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by liz155 View Post
    \int5x\sqrt[3]{1-x^2}dx

    I am required to use U subsitiution.

    u=  1-x^2
    du= -2x dx

    So the problem became....

    \frac{-2}{5} \int u^{1/3} du

    and my final answer was...

    \frac{-3}{10} u^{4/3} + C


    This seems correct to me, however the answer my book gives is...
    \frac{-15}{8} u^{4/3} + C

    Of course I will replace the u in the final answer, but I cannot figure out how the book did this problem. Can someone please tell me what I did wrong.

    Thanks so much!
    the constant you should have in front of the new integral is - \frac 52, try again
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  3. #3
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    How did you get -5/2?

    I thought that in order to replace the 5x dx with the du of -2x dx, I would have to multiply by some constant. I thought I had to get the 5x dx part equal to the du, but have a constant out in front. I multiplied by -2/5 so the 5's cancelled out and only -2 was left. I'm confused on how you get -5/2...

    Then solving the rest of the problem and reducing fractions led to the -3/10.

    Thanks for your help!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by liz155 View Post
    How did you get -5/2?

    I thought that in order to replace the 5x dx with the du of -2x dx, I would have to multiply by some constant. I thought I had to get the 5x dx part equal to the du, but have a constant out in front. I multiplied by -2/5 so the 5's cancelled out and only -2 was left. I'm confused on how you get -5/2...

    Then solving the rest of the problem and reducing fractions led to the -3/10.

    Thanks for your help!
    du = -2x~dx

    \Rightarrow - \frac 12 du = x~dx

    so our integral becomes:

    5 \cdot \left( - \frac 12 \right) \int u^{1/3}~du

    and continue...

    you should not seek to cancel the 5, it was a constant that was there, leave it alone. you should seek to cancel the -2 that you got from the substitution, so you multiply by -1/2
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  5. #5
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    Thanks soooo much!!! I see what I did wrong and after finishing the problem, I got the right answer!

    Thank you!!
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