A Can a function inside the integral be erased?

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

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.
 

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