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Math Help - Definite Integral

  1. #1
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    Definite Integral

    \int_{x=1}{x=0} (x^2(4x^3+2)^3dx

    u=4x^3+2
    du=12x^2(dx)
    \frac{du}{12}=x^2???????

    Lost on what to do with these problems after this.
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    Quote Originally Posted by Jim Marnell View Post
    \int_{x=1}{x=0} (x^2(4x^3+2)^3dx

    u=4x^3+2
    du=12x^2(dx)
    \frac{du}{12}=x^2???????

    Lost on what to do with these problems after this.
    Let u=4x^3+2\Rightarrow du=12x^2\,dx. Now we can substitute. And since this is a definite integral, we can leave things in terms of u by changing the limits of integration. When x=0,\;u=2, and when x=1,\;u=6. So,

    \int_1^0x^2\left(4x^3+2\right)^3\,dx

    =\int_1^0\left(4x^3+2\right)^3\left(x^2\,dx\right)

    =\frac1{12}\int_1^0\left(4x^3+2\right)^3\left(12x^  2\,dx\right)

    =\frac1{12}\int_6^2u^3\,du

    =-\frac1{12}\int_2^6u^3\,du.

    Can you finish?
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  3. #3
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    Thanks! yea ill try to finish this problem up and post what i get. Just wondering why 1/12 turns into -1/12 and the 2,6 switched to 6,2 and would the du become c?
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    Quote Originally Posted by Jim Marnell View Post
    Thanks! yea ill try to finish this problem up and post what i get. Just wondering why 1/12 turns into -1/12 and the 2,6 switched to 6,2...
    \int_a^bf(x)\,dx=-\int_b^af(x)\,dx.

    We usually evaluate definite integrals from left to right, so I switched the limits of integration. But you can apply the fundamental theorem of calculus without doing this; it works out the same way.

    ...and would the du become c?
    I do not understand what you mean here. Are you talking about the constant of integration? The du just indicates that the integral is to be evaluated with respect to u.
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  5. #5
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    ok both explanations make sense now, thanks again!
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  6. #6
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    would it become
    \frac{-1}{12}\times{\frac{u^4}{4}}+c

    then i would sub in the numbers for u then sub in the x=2 and x=6?
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    Quote Originally Posted by Jim Marnell View Post
    would it become
    \frac{-1}{12}\times{\frac{u^4}{4}}+c

    then i would sub in the numbers for u then sub in the x=2 and x=6?
    After changing the limits of integration, you do not need to back-substitute for u. Just evaluate it like normal:

    -\frac1{12}\int_2^6u^3\,du

    =-\frac1{12}\left[\frac{u^4}4\right]_2^6

    =-\frac1{12}\left[\frac{6^4}4-\frac{2^4}4\right]

    =-\frac1{12}(324-4)

    =-\frac{320}{12}=-\frac{80}3
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  8. #8
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    wow for all the problems like this ive been doing, i kept resubing in the u value. Thanks for clarifying this for me. i kept wondering why i was getting these strange numbers.
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  9. #9
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    Quote Originally Posted by Jim Marnell View Post
    wow for all the problems like this ive been doing, i kept resubing in the u value. Thanks for clarifying this for me. i kept wondering why i was getting these strange numbers.
    You can do it that way. For this problem, we would leave the limits of integration as is and get

    -\frac1{12}\int_{x=0}^{x=1}u^3\,du

    =-\frac1{12}\left[\frac{u^4}4\right]_{x=0}^{x=1}

    =-\frac1{12}\left[\frac{\left(4x^3+2\right)^4}4\right]_0^1

    =-\frac1{12}\left[\frac{\left(4\cdot1^3+2\right)^4}4-\frac{\left(4\cdot0^3+2\right)^4}4\right]

    =-\frac1{12}\left[\frac{6^4}4-\frac{2^4}4\right]

    =-\frac{320}{12}=-\frac{80}3.

    As you can see, the answer is the same.
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