How Do You Evaluate an Integral with a Function Transformation?

[armolinasf]

Homework Statement

suppose \int^{1}_{0} f(x)dx=k. Evaluate \int^{1}_{0} xf(1-x^{2}[/tex). Give your answer in terms of k.<br /> <br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2> <br /> <br /> I&#039;m not too sure how to go about solving this one. I think there&#039;s some substitution involved but like I said I&#039;m not sure. Thanks for the help

 

[Dick]

u=1-x^2 looks like a good bet. Try it.

 

[armolinasf]

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

But where would I go from there?

 

[S_klogW]

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.

 

[armolinasf]

How would I express that in terms of k?

 

[SammyS]

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

 

[armolinasf]

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?

 

[armolinasf]

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

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

 

[Dick]

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

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

You've maybe partially got it. But that's not quite right. Don't forget to change the limits on the u integral.