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Thread: Definite Integral

  1. #1
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    Definite Integral

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

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

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

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

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

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

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

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

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

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

    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...
    $\displaystyle \int_a^bf(x)\,dx=-\int_b^af(x)\,dx.$

    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 $\displaystyle du$ just indicates that the integral is to be evaluated with respect to $\displaystyle u.$
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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
    $\displaystyle \frac{-1}{12}\times{\frac{u^4}{4}}+c$

    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
    $\displaystyle \frac{-1}{12}\times{\frac{u^4}{4}}+c$

    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 $\displaystyle u.$ Just evaluate it like normal:

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

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

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

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

    $\displaystyle =-\frac{320}{12}=-\frac{80}3$
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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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    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

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

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

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

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

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

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

    As you can see, the answer is the same.
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