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Math Help - definate integral question

  1. #1
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    definate integral question

    between 0 and t

    x^2e^(-x^3) dx

    Since t is not determined, I'm guessing the second part of the integral is just going to be variables, but I'm having some difficulty getting there.

    For likely the same reasons:

    between 0 and 2pi (theta)sin(theta)^2 d(theta) is a bugger. halp?
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  2. #2
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    For the first integral, try the substitution  u = -x^3 .

    For the second one, try integration by parts, using the reduction formulas for  \int sin^2(\theta) d\theta and  \int cos^2(\theta) d\theta .
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  3. #3
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    Quote Originally Posted by JG89 View Post
    For the first integral, try the substitution  u = -x^3 .

    For the second one, try integration by parts, using the reduction formulas for  \int sin^2(\theta) d\theta and  \int cos^2(\theta) d\theta .
    We're not into integration by parts, but what about substituting -x^3 for u? Can you describe what I would have to do for that method?
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  4. #4
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    If you let  u =- x^3 then it follows that  \frac{-du}{3} = x^2 .
    For the second integral, is it  \int \theta sin^2(\theta) d\theta or  \int \theta sin(\theta^2)d\theta ? If itís the latter, the substitution  u = \theta^2 should do the trick.
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  5. #5
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    So by the power rule for integration I got:

    (u^(n+1)) / (n+1) = u^2/2 and therefore the integration of sin is -cos, plug that into u^2/2 and you get (-cosx^2)/2 (x is easier than theta to show)

    except I'm not sure where theta in front of the sin in the original problem and since it is between 0 and 2pi, how to get only the area in that range. Sorry if the questions sound dumb, but in this summer class we do a chapter in a day, so there was 5 min spent on this, and I'm afraid it didn't sink in...
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  6. #6
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    I found it hard to understand what you are trying to say, but I think I got the gist of it.

    We'll use x instead of theta. You're right that an antiderivative of the given integral is  F(x) = \frac{-cos(x^2)}{2} . To evaluate the area between x = 0 and x = 2pi, then you have to calculate  F(2\pi) - F(0) .
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  7. #7
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    You can use substitution in both cases.

    In the first integral, let u=-x^3

    so du=-3x^2dx

    So the integral reduces to (-e^u)/3 du
    with the limits ranging from 0 to -t^3

    Hence the answer would be -1/3(e^(t^3) - 1)
    =1/3{1-e^(t^3)}
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