Find a function $f$ such that $\int_0^{P(x)} f(t) dt = 1- e^{2P(x)}$

Question

I'm trying to solve the following homework problem. It states as follows:

"Let $P(x)$ be a polynomial such that $P'(x) \neq 0$ for all values of $x$. Does there exist a continuous function $f$ such that $$ \int_0^{P(x)} f(t) dt = 1- e^{2P(x)} \quad ? $$ If it does exist, show one of these functions. If not, prove that they don't exist".

My attempt at a solution: I differentiated the equation given to try to get rid of the integral. I get to the equation $f(P(x)) = -2e^{2P(x)} P'(x)$. From here I tried to see if there is a way to express $P'(x)$ in term of $P(x)$, or in other words, I tried to find a function $g$ such that (for any polynomial) $g(P(x)) = P'(x)$. If I where to find this, then the function $f$ could be written as $f(x) = -2e^{2x} g(x)$, but I don't know if this is possible.

Am I on the right path? If anyone has any suggestions on how to go about this problem it would be very helpful. Thank you!

Answer 1

You had the right idea (use the $2^{nd}$ FTC to differentiate both sides) but didn't do it quite right. The proper way: if $f(t)$ is continuous and $b(x), a(x)$ are differentiable, $$ \frac{d}{dx}\int _{a(x)}^{b(x)}f(t)\,dt = f(b(x))b'(x)- f(a(x))a'(x) $$Multiplication of the derivatives comes from the Chain Rule. So, differentiating we have $$ f(P(x))P'(x) = -2P'(x) e^{2P(x)} $$Since $P'(x)\neq 0$ for any $x$, we can cancel these terms: $$ f(P(x)) = -2 e^{2P(x)} $$Indeed, if we have $f(t)=-2e^{2t}$, then the condition is satisfied.

Answer 2

If you derive the equality you get $$ P'(x) f(P(x)) = -2 P'(x)e^{2 P(x)} \Leftrightarrow P'(x)\left( f(P(x))+2e^{2P(x)}\right)=0. $$

So, $f(t)=-2 e^{2t}$ is your candidate. Now you must verify that it actually works...

$$ \int_0^{P(x)} -2 e^{2t}dt = \left[-e^{2t}\right]_0^{P(x)} = 1-e^{2 P(x)}. $$

At this point we can observe that we can drop the hypothesis $P'(x)\ne 0$.