# Integral help

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

HallsofIvy
Homework Helper
Since the integral involves g', start with integration by parts: let u= f, dv= g'dx. Then du= f'dx and v= g.
$$\int_0^{10} f(x)g'(x)dx= f(x)g(x)\left|_{x=0}^{10}- \int_0^{10}f'(x)g(x)dx$$
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
$$670- \int_0^{10}2xg(x)dx$$
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.

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!!

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.

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

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