# How would I solve this integral?

suppose $$\int^{1}_{0}$$ f(x)dx=k. Evaluate $$\int^{1}_{0}$$ xf(1-x$$^{2}[/tex). Give your answer in terms of k. ## The Attempt at a Solution I'm not too sure how to go about solving this one. I think there's some substitution involved but like I said I'm not sure. Thanks for the help ## Answers and Replies Dick Science Advisor Homework Helper u=1-x^2 looks like a good bet. Try it. That would give me [tex]\int^{1}_{0}$$$$\sqrt{1-u}$$f(u)

But where would I go from there?

By the way, I think your question is to calculate: ∫（from 0 to 1）xf(1-x^2)dx.
just let u=1-x^2, so xdx=d(0.5x^2)=-0.5du
so the integral is solved.

How would I express that in terms of k?

SammyS
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That would give me
$$\int^{1}_{0}$$$$\sqrt{1-u}$$f(u)

But where would I go from there?
You forgot the du.

$$u=1-x^2\quad\to\quad du=-2x\,dx$$

So, $$xf(1-x^2)\,dx=-(1/2)f(1-x^2)\,(-2x)\,dx=f(u)\,du$$

How about this: since Int (from 0 to 1) f(x)dx=k and Int (from 0 to 1) xf(1-x^2) when we let u=1-x^2 becomes Int (from 0 to 1) sqrt(1-u)f(u), wouldn't that be saying sqrt(1-u)*Int (from 0 to 1) f(u), which is no different then Int (from 0 to 1) f(x)dx=k. So the answer would be sqrt(1-u)*k, or sqrt(1-(1-x^2))*k= sqrt(x^2)*k = xk?

So then including the -1/2 it would be -xk/2

Edit: I understand this now: I got (-1/2)k as an answer

Last edited:
Dick