Can a function inside the integral be erased?

[CECE2]

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 $$\int_a^b g(x) \, dx = \int_a^b h(x) \, dx$$?

 

[Drakkith]

Hi CECE2. For ease of reading, would you mind putting this into LaTeX format? You can find the LaTeX help section here: https://www.physicsforums.com/help/latexhelp/

 

[Mark44]

The OP has reformatted the original post.

 

[pasmith]

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 $$\int_a^b g(x) \, dx = \int_a^b h(x) \, dx$$?

So you have $$\int_a^b f(x)g(x)\,dx - \int_a^b f(x)h(x)\,dx =<br /> \int_a^b f(x)(g(x) - h(x))\,dx = \int_a^b F(x)\,dx = 0.$$ You cannot in general conclude from $\int_a^b F(x)\,dx = 0$ that $F(x) \equiv 0$ everywhere on $(a,b)$. You can only reach this conclusion if you know in addition that $F(x) \geq 0$ everywhere or that $F(x) \leq 0$ everywhere.

 

[Mark44]

Thread locked. The OP started a new thread on the same question here https://www.physicsforums.com/threads/can-a-function-inside-the-integral-be-erased.1060518/.