Can a function inside the integral be erased?

  • #1
CECE2
5
1
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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  • #3
The OP has reformatted the original post.
 
  • #4
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 [tex]\int_a^b f(x)g(x)\,dx - \int_a^b f(x)h(x)\,dx =
\int_a^b f(x)(g(x) - h(x))\,dx = \int_a^b F(x)\,dx = 0.[/tex] You cannot in general conclude from [itex]\int_a^b F(x)\,dx = 0[/itex] that [itex]F(x) \equiv 0[/itex] everywhere on [itex](a,b)[/itex]. You can only reach this conclusion if you know in addition that [itex]F(x) \geq 0[/itex] everywhere or that [itex]F(x) \leq 0[/itex] everywhere.
 

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