Question

given that F'(x)=f(x), the value of ∫(a,b) x f(x^2)dx is...
is it (b^2F(b^2)-a^2F(a^2))/2 by the fundamental theorem part II?

Answer 1

hard to say .... can you clean that up any?

Answer 2

im thinking you would want the derivative of a and b ... thanks to the chain rule

Answer 3

well my thought process was that you separate integrals getting ∫x +∫f(x^2) and anti derive: that would give you (a^2/2)-(b^2/2) but i wasn't sure what to do next

Answer 4

because i don't know what to do with the f(x^2)

Answer 5

you cant seperate a product, only a sum

Answer 6

oh okay

Answer 7

well, you can work a product in a few ways.\[\int u~dv=uv-\int v~du\] is one method
but im still unsure how to read all the info from the post.

Answer 8

another possibility is
\[\int_{a}^{b} F'(g(u))~g'(u)~du = F(g(b))-F(g(a)) \]

Answer 9

since x^2 pops out a 2x, that seems to be a more reasonable setup to me

Answer 10

well the options for the answer are
A. (F(b^2)-F(a^2))/2
B. bF(b^2)-aF(a^2)
C. 2F(√b)-2F(√a)
D. (b^2F(b^2)-a^2F(a^2))/2
E. 2F(b^2)-2F(a^2)
so I'm not really sure how to get any of those

Answer 11

@amistre64 is on the right track. Recognize it's equal to \(\frac12\int_a^b 2x F(x^2)\mathrm{d}x\).

Answer 12

oops, I messed up the case of \(f\).$$\int_a^bxf(x^2)\mathrm{d}x=\frac12\int_a^b2xf(x^2)\mathrm{d}x=\frac12\int_{a^2}^{b^2}F(x)\mathrm{d}x=\frac{F(b^2)-F(a^2)}{2}$$

Answer 13

wonderful, thank you both!

Answer 14

Sorry I screwed up the casing again. The third integrand is supposed to have \(f(x)\) not \(F(x)\).

Answer 15

you did well :)

Answer 16

\[\int_a^bxF'(x^2)dx\]let x^2 = u
at x=a, u=a^2
at x=b, u=b^2
\[\frac12\int_{a^2}^{b^2}F'(u)du=\frac12(F(b^2)-F(a^2))\]
is prolly how i would have notated it, same same tho