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Homework Help: Integration by parts problem

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data

    Estimate [tex] \int_{0}^{10} f(x) g'(x) dx [/tex] for f(x) = [tex]x^{2}[/tex]
    and g has the values in the following table.

    \begin{array}{l | c|c|c|c|c|c |}
    \hline g&0&2&4&6&8&10\\
    \hline g(x)&2.3&3.1&4.1&5.5&5.9&6.1\\

    2. Relevant equations

    [tex] \int uv' dx = uv = \int u'v dx [/tex]

    3. The attempt at a solution

    Okay so, since f(x) is x squared i chose

    u = [tex]x^{2}[/tex] and v' = g'(x)
    u' = 2x dx and v = g(x)

    plugging in...

    [tex]g(x)x^{2} - \int_{0}^{10} 2xg(x) dx [/tex]

    and this is where im stuck. I cant plug in the g values because i first need to take the integral of 2xg(x) ....I think. lol

    a nudge in the right direction would be ub3r helpful and much appreciated. thanks!

    ps. that latex table took me like a half hour to figure out rofl :biggrin:
  2. jcsd
  3. Sep 29, 2008 #2


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    Well, the integral

    [tex]\int_0^{10} 2xg(x)dx[/tex]

    gives the area under the curve [itex]2xg(x)[/itex] between [itex]x=0[/itex] and [itex]x=10[/itex].

    You are given g(x)at certain points along the interval, so what is 2xg(x) at those points? Draw a picture and see if you can find a way to estimate the area under 2xg(x):wink:
  4. Sep 29, 2008 #3
    [tex]\int_0^{10} 2xg(x)dx[/tex]

    when x= 0, 2xg(x) = 0
    x=2, 2xg(x) = 2(2)(3.1) = 12.4
    x=4, 2xg(x) = ... = 32.8
    x=6, 2xg(x) = ... = 66
    x=8, 2xg(x) = ... = 94.4
    x=10, 2xg(x) = ... = 122

    connect thesse and estimate area under from 0 to 10?
    makes sense, but is there any other way to solve the problem?
  5. Sep 29, 2008 #4


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    Because you are only given a few values of g(x) and not an explicit functional no exact solution will be possible.

    You might be able to get a slightly more accurate value by fitting a 4th degree polynomial to the points you are given, but it will still just be an approximation and I don't think your instructor is looking for anything that complicated.
  6. Sep 29, 2008 #5
    okay awesome!! thanks for helping me out!
  7. Sep 29, 2008 #6
    oh wait how do i work in that [tex]g(x)x^{2} [/tex] part
  8. Sep 29, 2008 #7


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    Gold Member

    You mean


    Remember, integration by parts means that [itex]uv[/itex] is evaluated at the endpoints of your integration interval.
  9. Sep 29, 2008 #8
    geez this problem is pwning me lol.
    so we go...

    [tex]g(x)x^2 - \int_0^{10} 2xg(x)dx[/tex]
    (2.3)(0) - 2(0)(2.3) = 0, for x=0
    (3.1)(4) - 2(2)(3.1) = 0, for x=2
    (4.1)(16) - 2(4)(4.1) = 54.4, for x=4
    (5.5)(36) - 2(6)(5.5) = 132, for x=6
    ...and so on...

    calculate area under (connected) points (0,0) (0,0) (4, 54.4) (6, 132) .......?
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