Can a function inside the integral be erased?

Given that $$\int_a^b f(x)g(x) \, dx = \int_a^b f(x)h(x) \, dx$$ and $$f(x)=e^x$$, is it true that $$\int_a^b g(x) \, dx = \int_a^b h(x) \, dx$$?

Does the first equation hold for any ##a## and ##b## or for are ##a## and ##b## fixed?

Hill said:

Does the first equation hold for any ##a## and ##b## or for are ##a## and ##b## fixed?

##a## and ##b## are fixed scalar values

CECE2 said:

##a## and ##b## are fixed scalar values

Then my answer is, no, not necessarily. ##g(x)## and ##h(x)## can "cut out" different sections of the ##f(x)##.

For your consideration... the laziest, least rigorous, counter example; graphic novel style.

Hill said:

Then my answer is, no, not necessarily. ##g(x)## and ##h(x)## can "cut out" different sections of the ##f(x)##.

Thank you

DaveE said:

For your consideration... the laziest, least rigorous, counter example; graphic novel style.
View attachment 341189

Thank you! one counter example must be enough

CECE2 said:

Given that $$\int_a^b f(x)g(x) \, dx = \int_a^b f(x)h(x) \, dx$$ and $$f(x)=e^x$$, is it true that $$\int_a^b g(x) \, dx = \int_a^b h(x) \, dx$$?

Note that the question is equivalent to this. Given that
$$\int_a^b e^xf(x) \ dx = 0$$Is it true that:
$$\int_a^bf(x) \ dx = 0$$

And if we let ##f(x) = e^{-x}g(x)## above, then the question is equivalent to:

Given that
$$\int_a^b g(x) \ dx = 0$$Is it true that:
$$\int_a^be^{-x}g(x) \ dx = 0$$Which doesn't look very likely. In fact, in that formulation it's fairly clear that the set of functions that fail the original hypothesis must bean infinite dimensional subspace of the space of integrable functions.