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Math Help - Integration tables

  1. #1
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    Integration tables

    I was given the following problem:

    A. Use Derive to evaluate the integral \int x^5*(x^2+1)^{1/2} dx

    B. Use integration tables in the back of your book to evaluate the integral from part A. State the table entry used and show all work. After you find this antiderivative by hand, show that it is equal to the antiderivative given by derive.

    While using derive, I obtained this as the answer for part A:

    = \frac{(x^2+1)^{1/2}(15x^6+3x^4-4x^2+8)}{105}

    But using my Ti-89 I got this instead:

    = \frac{(x^2+1)^{3/2}(15x^4-12x^2+8)}{105}

    So I'm not sure what's going on there, but I went ahead and tried out an integration table that looked useful:

    \int u^n(bu+a)^{1/2} du = \frac{2}{b(2n+3)}(u^n(a+bu)^{3/2} - na \int u^{n-1}(bu+a)^{1/2} du

    However, when I use u=x, n=5 and b=x the integral at the end looks just as messy as what I started with. Do I really have to use the integration table again on the last integral part of the integration table? Am I doing it wrong?

    Edit: I found an online pdf of the integration tables I have access to at this site: http://teachers.sduhsd.k12.ca.us/abr...lesStewart.pdf
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ebonyscythe View Post
    = \frac{(x^2+1)^{1/2}(15x^6+3x^4-4x^2+8)}{105}

    = \frac{(x^2+1)^{3/2}(15x^4-12x^2+8)}{105}
    Note that (x^2 + 1)^{3/2} = (x^2 + 1)(x^2 + 1)^{1/2}. So what is
    (x^2 + 1)(15x^4 - 12x^2 + 8)

    -Dan
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  3. #3
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    Ah, so they are equal. I didn't look at it that way at all. Thanks for that, now I can put that part aside.

    Can anyone help with the integration tables next?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ebonyscythe View Post
    So I'm not sure what's going on there, but I went ahead and tried out an integration table that looked useful:

    \int u^n(bu+a)^{1/2} du = \frac{2}{b(2n+3)}(u^n(a+bu)^{3/2} - na \int u^{n-1}(bu+a)^{1/2} du

    However, when I use u=x, n=5 and b=x the integral at the end looks just as messy as what I started with. Do I really have to use the integration table again on the last integral part of the integration table? Am I doing it wrong?

    Edit: I found an online pdf of the integration tables I have access to at this site: http://teachers.sduhsd.k12.ca.us/abr...lesStewart.pdf
    \int x^5*(x^2+1)^{1/2} dx

    Let u = x^2. Then your integral becomes
     = \int u^{5/2}(u + 1)^{1/2}~\frac{1}{2 u^{1/2}}~du

    = \frac{1}{2} \int u^2(u + 1)^{1/2}~du

    So n = 2, b = 1, and a = 1.

    Thus according to the table:
    \int u^2(u + 1)^{1/2}~du = \frac{2}{7} \left ( u^2(1+u)^{3/2} - 2 \int u(u+1)^{1/2}~du \right )

    and
    \int u(u+1)^{1/2}~du = \frac{2}{5} \left ( u(1+u)^{3/2} - \int (u+1)^{1/2} ~du \right )

    and
    \int (u+1)^{1/2} ~du = \frac{2}{3}(u + 1)^{3/2}

    So
    \int u^2(u + 1)^{1/2}~du = \frac{22}{105}u^2(1+u)^{3/2}
    after a lot of simplifying.

    Thus
    \int x^5(x^2+1)^{1/2}~dx = \frac{11}{105}x^4(x^2 + 1)^{3/2}

    Which is obviously not your answer. Unless I missed some factors something is wrong in the table.

    -Dan
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  5. #5
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    I see what I did wrong. I didn't perform the U-substitution right away, like you did, and must've made it way harder on myself. Thank you so much for the help, I think that's exactly what I need... now to absorb it.
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  6. #6
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    Quote Originally Posted by topsquark View Post
    = \frac{1}{2} \int u^2(u + 1)^{1/2}~du

    So n = 2, b = 1, and a = 1.

    Thus according to the table:
    \int u^2(u + 1)^{1/2}~du = \frac{2}{7} \left ( u^2(1+u)^{3/2} - 2 \int u(u+1)^{1/2}~du \right )
    Question, what happened to the 1/2 that was in front of the first integral?
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ebonyscythe View Post
    Question, what happened to the 1/2 that was in front of the first integral?
    The factor of 1/2 relates the x integral to the u integral.

    -Dan
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