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Math Help - integration by substitution

  1. #1
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    integration by substitution




    (a) Let u=x^4 then du=4x^3dx and therefore \frac{du}{4x^3}=dx. We get:

    \int^b_{a} f(x^4)dx= \int^{1^4}_{(-1)^4} 1+u \frac{du}{4x^3} =\int^{1}_{1} 1+u \frac{du}{4x^3}

    Both limits of integration in the new integral are 1, so the value of the definite integral is 0.

    (b)
    \frac{du}{dx}=4x^3

    It is only zero when x=0. Jacobian become zero at (0,0) so it vanishes at this point.

    (c) This is the part I'm really stuck on: from part (a) I know what to substitute for "dx" when I make the substitution u=x^4. So I think this is what the question is asking for:

    \int^0_{-1} 1+u \frac{du}{4x^3} + \int^1_0 1+u \frac{du}{4x^3}

    But expressions involving x should not appear in the new integrals. How do I get rid of the " 4x^3"? Can anyone show me how to do this part?
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  2. #2
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    Since u=x^4 is the substitution being made, you can take a fourth root on both sides to get x=\pm \sqrt[4]{u}. The \pm is important here, and is why the integral needs to be split into two regions.

    In the region -1<x<0, we should get x=-\sqrt[4]{u} because x is always negative in this region. Using the same logic, for 0<x<1, we should use x=\sqrt[4]{u}. Now that you have an expression for x in both regions you can make the substitution.

    By the way, make sure you use parentheses in your integral. It should be like this:

    \int^0_{-1} (1+u) \frac{du}{4x^3} + \int^1_0 (1+u) \frac{du}{4x^3}
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  3. #3
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    I'm pretty sure that you need to set the limits also. For each of your integrals you approach a value that is undefined.
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