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Math Help - Definite integral problem

  1. #1
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    Definite integral problem

    The question:

    Find \int_{0}^{1} 2x(1 + x^2)^3 dx

    My attempt:
    Let u = 1 + x^2

    \frac{du}{dx} = 2x

    du = 2x dx

    \int_{0}^{1} u^3 du =  _{0}^{1}[3u^2]

    Find bounds in terms of u...
    Sub 1 into u:
    1 + (1)^2 = 2
    Sub 0 into u:
    1 + 0 = 1

    _{1}^{2}[3u^2]
    [3(2)^2] - [3(1)^2] = 9

    However, the answer is 15/4. :/

    What am I doing wrong? Thanks!
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  2. #2
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    What is the derivative of \dfrac{\left(1+x^2\right)^4}{4}~?
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  3. #3
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    2x(1+x^2)^3
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  4. #4
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    Well there you have it. Do you not?
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  5. #5
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    Not really, because you've solved the integral and I just proved it. I can't always make that leap like you did, which is why I tried the substitution method to help me. I'm still not sure what I've actually done wrong.
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  6. #6
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    Quote Originally Posted by Glitch View Post
    The question:

    Find \int_{0}^{1} 2x(1 + x^2)^3 dx

    My attempt:
    Let u = 1 + x^2

    \frac{du}{dx} = 2x

    du = 2x dx

    \int_{0}^{1} u^3 du =  _{0}^{1}[3u^2] <=== This is the derivative not the integral which should be \frac14 u^4

    Find bounds in terms of u...
    Sub 1 into u:
    1 + (1)^2 = 2
    Sub 0 into u:
    1 + 0 = 1

    _{1}^{2}[3u^2]
    [3(2)^2] - [3(1)^2] = 9

    However, the answer is 15/4. :/

    What am I doing wrong? Thanks!
    ...
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  7. #7
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    Nevermind, I was finding the derivative of u^3 instead of the anti-derivative like an idiot. All solved.

    EDIT: Yeah, just realised earboth. Thanks.
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