Integration using the fundamental theorem of calculus

1. Nov 7, 2013

trulyfalse

Hello PF.

1. The problem statement, all variables and given/known data
Find a function g such that

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

2. Relevant equations

From the fundamental theorem of calculus:

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

3. 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? :)

2. Nov 7, 2013

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).

3. Nov 7, 2013

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)?

4. Nov 7, 2013

brmath

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 $x^2$.

5. Nov 7, 2013

trulyfalse

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

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