Integration using the fundamental theorem of calculus

[trulyfalse]

Hello PF.

Homework Statement

Find a function g such that

$\int_0^{x^2} \ tg(t) \, \mathrm{d}t = x^2+x$

Homework Equations

From the fundamental theorem of calculus:

$f(x) = \frac{d}{dx}\int_a^x \ f(t) \, \mathrm{d}t$

The Attempt at a Solution

After taking the derivative of both sides of the equation:

$\frac{d}{dx}\int_0^{x^2} \ tg(t) \, \mathrm{d}t = 2x+1$

Thus, from the fundamental theorem and chain rule,

$(2x)(x^2)g(x^2) = 2x+1$

$2x^3g(x^2) = 2x+1$

$g(x^2) = \frac{1}{x^2} + \frac{1}{2x^3}$


However, I know that this answer is wrong because the definite integral of this function (when substituted into the original equation) is not equal to x + x^2. I'm having difficulty identifying my error. Could someone please point me in the right direction? :)

 

[brmath]

What did you put back into your integral for checking. You solved only down to g($x^2$). Did you put that back in or g(x)? I got the right answer when I put in g(x).

 

[trulyfalse]

I put g(x^2) back into the integral. How would I simplify g(x^2) to get g(x)? Would I square every expression for x in the equation (i.e x^2 and x^3)?

 

[brmath]

I put g(x^2) back into the integral. How would I simplify g(x^2) to get g(x)? Would I square every expression for x in the equation (i.e x^2 and x^3)?

To break it into two steps: Let u = $x^2$ and figure out what g(u) is. Then since it doesn't matter what symbol you use the function g(u) is the same as g(x).

More directly, just stick an x in everywhere you have an $x^2$.

 

[trulyfalse]

Aha! Then U = sqr(x) and that can be substituted in the equation. Thanks brmath!