Thread: Fundmental Theorem of Calculus problem #2

1. Fundmental Theorem of Calculus problem #2

The question is: The function f is differentiable on (−∞, ∞). Determine f (2) in the following cases:

$\int_0^{f(x)}{t^2}dx={x^2}{(1+x)}}$

So I let ${u}={f(x)}$ ==> $\int_0^u{t^2}{dt}$ ==>
$\frac{d}{du} \int_0^u{t^2}{dt}={(f(u))^2}$

So $\frac{d}{dx} \int_0^u {t^2}{dt}={(f(u))^2} \frac{du}{dx}$ ==> ${(f(x))^2}{f'(x)}={2x}+{3x^2}$

Assuming that my steps are correct (and I'm not certain that they are), where do I go from here? For instance, what is ${f'(x)}$?

Many thanks for any suggestions for this problem!
CP

2. Re: Fundmental Theorem of Calculus problem #2

First, I suspect there is a typo in your statement of the problem. Your integrand is [tex]x^2[tex] but the integral with respect to x!
I suspect you meant $\int_0^{x^2} t^2 dt$. Is that correct? Assuming it is-

I'm not sure why you would differentiate. If you just go ahead and integrate $\int t^2 dt= \frac{1}{3}t^3+ C$ so that $\int_0^{f(x)} t^2 dt= \frac{1}{3} (f(x))^3$.

So your equation is $\frac{1}{3}(f(x))^3= x^2(1+ x)$

Solve that for f(x). (f(x) is just a number. This is exactly the same as solving $\frac{1}{3}y^3= x^2(1+ x)$ for x.)