Integral help

  • Thread starter blumfeld0
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  • #1
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Find the integral of:
f(x)g'(x)dx
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

Thanks
 

Answers and Replies

  • #2
HallsofIvy
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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.
 
  • #3
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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?

thanks!!
 
  • #4
590
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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.
 
  • #5
148
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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:

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