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Integral help

  1. Sep 26, 2006 #1
    Find the integral of:
    from zero to ten.
    If f(x) = x^2 and g has the following values on the table

    at x=0, g(X)=2
    at x=2, g(x)=2.7
    at x=4, g(x)=3.8
    at x=6, g(x)=4.6
    at x=8, g(x)=6.0
    at x=10, g(x)=6.7

    I know that I have to approximate the integral by finding the average of the left and right sums

    I just need help getting started. Any help would be appreciated

  2. jcsd
  3. Sep 26, 2006 #2


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    Science Advisor

    Since the integral involves g', start with integration by parts: let u= f, dv= g'dx. Then du= f'dx and v= g.
    [tex]\int_0^{10} f(x)g'(x)dx= f(x)g(x)\left|_{x=0}^{10}- \int_0^{10}f'(x)g(x)dx[/tex]
    You know that f(0)= 0, g(0)= 2, f(10)= 100, g(10)= 6.7 so that first term is 670. Also f(x)= x2, f'(x)= 2x so your integral is
    [tex]670- \int_0^{10}2xg(x)dx[/tex]
    Since you know g(x) at x= 0, 2, 4, 6, 8, 10, you can calculate 2xg(x) at x= 0, 2, 4, 6, 8, 10 and then do a numerical approximation.
  4. Sep 26, 2006 #3
    omg. thanks. thats genius! ok so i sat down and got up to what you got. in addition i know
    x=0, 2*x*g(x)= 0
    x=2, 2*x*g(x)= 10.8
    x=4, 2*x*g(x)= 30.4
    x=6, 2*x*g(x)= 55.2
    x=8, 2*x*g(x)= 96
    x=10, 2*x*g(x)= 134

    so the integral from zero to ten of 2*x*g(x)= ??

    and i still dont understand how i calculate the left and right sums?

  5. Sep 26, 2006 #4
    Hint: Definite integral of a function wrt x can be seen as the area bounded by the curve and the x axis. You can approximately calculate the area using rectangles whose dimensions are obtained from the values you have shown in your last post.
  6. Sep 26, 2006 #5
    Yes the integral is the area under the curve from zero to ten
    SO, is it accurate?
    to say that the
    left sum = (0*2) + (10.8*2)+(30.4*2)+(55.2*2)+(96*2)= 384.8??
    right sum = (10.8*2)+(30.4*2)+(55.2*2)+(96*2)+ (134*2)= 652.8??
    average of left sum and right sum= 518.8??

    so my final answer is 670-1054.8 = 151.2??

    thank you
    Last edited: Sep 26, 2006
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