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Math Help - integration help

  1. #1
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    integration help

    I have to integrate

    (t+1)(t^2+4)

    from -3^1/2 to +3^1/2

    I know the answer comes out to 10(3)^1/2, but I just can't get it. I tried re-writing it as the integral of t^3+t^2+4t+4 and that just made everything a mess. I also tried to do the integral of (t+1) time the integral of (t^2+4). However, I'm not sure if you can even do that. Can anyone help me with this problem?

    Also, this one is giving me trouble (substitution). I'm not sure if I did it right..


    Evaluate

    10x^1/2
    --------------dx
    (1+x^3/2)^2

    So I made

    u=1+x^3/2
    du=3/2x^1/2

    So then I know that 20/3 will get me what I need.. so..

    20/3 integral u^-2

    20/3 u^-1/-1

    Put it back in the original:

    20/3 ((10x^1/2)(1+x^3/2)^-1
    --------------------------
    -1

    Am I on the right track?
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  2. #2
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    Quote Originally Posted by joseph_ View Post
    I have to integrate

    (t+1)(t^2+4)

    from -3^1/2 to +3^1/2

    I know the answer comes out to 10(3)^1/2, but I just can't get it. I tried re-writing it as the integral of t^3+t^2+4t+4 and that just made everything a mess. I also tried to do the integral of (t+1) time the integral of (t^2+4). However, I'm not sure if you can even do that. Can anyone help me with this problem?

    Also, this one is giving me trouble (substitution). I'm not sure if I did it right..


    Evaluate

    10x^1/2
    --------------dx
    (1+x^3/2)^2

    So I made

    u=1+x^3/2
    du=3/2x^1/2

    So then I know that 20/3 will get me what I need.. so..

    20/3 integral u^-2

    20/3 u^-1/-1

    Put it back in the original:

    20/3 ((10x^1/2)(1+x^3/2)^-1 (*)
    --------------------------
    -1

    Am I on the right track?
    For the first one, you're on the right track - just multiple out and integral term by term. In general

    \int fg \, dx \ne \int g dx \int g dx

    For the second, everything looks good (I'm guessing the -1 underneath should be in front) except for the last step (see *), why put in the 10x^{1/2}?
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  3. #3
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    Okay I got the first one.

    On the second one:



    20/3 ((10x^1/2)(1+x^3/2)^-1 (*)
    --------------------------
    -1
    Don't I have to put the 10x^1/2 back in? Like you substitute out something integrate with respect to u then put it back in the original?

    And the -1 (I thought) belonged there. Now I'm just confused

    The -1 has to be on the bottom so it can get rid of the -1 when it comes down?

    and to be more clear this is what I have

    20 ((10x^1/2)(1+x^3/2)^-1 (*)
    --------------------------
    (3) -1

    ?
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  4. #4
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    Quote Originally Posted by joseph_ View Post
    Okay I got the first one.

    On the second one:



    20/3 ((10x^1/2)(1+x^3/2)^-1 (*)
    --------------------------
    -1
    Don't I have to put the 10x^1/2 back in? Like you substitute out something integrate with respect to u then put it back in the original?

    And the -1 (I thought) belonged there. Now I'm just confused

    The -1 has to be on the bottom so it can get rid of the -1 when it comes down?

    and to be more clear this is what I have

    20 ((10x^1/2)(1+x^3/2)^-1 (*)
    --------------------------
    (3) -1

    ?
    First \frac{1}{-1} = -1. Second, after you integrate, you have

    -\frac{20}{3} \frac{1}{u} = - \frac{20}{3} \frac{1}{1+x^{3/2}} - you substitute just where you see a u.
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  5. #5
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    Quote Originally Posted by danny arrigo View Post
    First \frac{1}{-1} = -1. Second, after you integrate, you have

    -\frac{20}{3} \frac{1}{u} = - \frac{20}{3} \frac{1}{1+x^{3/2}} - you substitute just where you see a u.
    So I'm doing my substitution incorrectly.

    Up to

    20/3 integral u^-2

    I integrate that.

    20/3 u^-1

    So I plug back in 1+x^3/2 into where I see u, then put a one over it and move it to the denominator.

    which gives me:


    20 (1)
    -----------------
    3 (1+x^3/2)

    So now I'm just confused where you got the negative from?
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  6. #6
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    Quote Originally Posted by joseph_ View Post
    So I'm doing my substitution incorrectly.

    Up to

    20/3 integral u^-2

    I integrate that.

    20/3 u^-1

    So I plug back in 1+x^3/2 into where I see u, then put a one over it and move it to the denominator.

    which gives me:


    20 (1)
    -----------------
    3 (1+x^3/2)

    So now I'm just confused where you got the negative from?
     \frac{20}{3} \int \frac{du}{u^2} = \frac{20}{3} \int u^{-2} \,du = \frac{20}{3} \frac{u^{-1}}{-1} + c = - \frac{20}{3} \frac{1}{u} + c - you see it now?
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  7. #7
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    Quote Originally Posted by danny arrigo View Post
     \frac{20}{3} \int \frac{du}{u^2} = \frac{20}{3} \int u^{-2} \,du = \frac{20}{3} \frac{u^{-1}}{-1} + c = - \frac{20}{3} \frac{1}{u} + c - you see it now?
    Yes from the bottom. Thanks.

    I have one more question

    I have a problem where it wants me to find average value (where you integrate then divide by the length, in this case 1)

    h(x) = -|x| on -1, 0 and a few others.

    I'm just not sure how to do the integral of that. If I make it 1/2x^2 I won't get the abs value... not sure where to go on this?
    Last edited by joseph_; February 25th 2009 at 09:02 AM. Reason: error.
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