Can a function inside the integral be erased?

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Discussion Overview

The discussion revolves around the implications of an integral equality involving a function multiplied by two different functions, specifically examining whether the functions can be equated under certain conditions. The scope includes mathematical reasoning related to integrals and functions.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • The original post (OP) poses a question about whether the equality of two integrals implies the equality of the integrals of the functions being multiplied by a common function.
  • Some participants suggest that the transformation of the integrals leads to an expression that equals zero, but caution that this does not necessarily imply that the function involved is zero everywhere on the interval.
  • There is a mention of needing additional conditions, such as the function being non-negative or non-positive, to conclude that the function must be zero everywhere.

Areas of Agreement / Disagreement

Participants express differing views on whether the conclusion about the functions can be drawn from the integral equality, indicating that the discussion remains unresolved.

Contextual Notes

Participants note limitations regarding the assumptions needed to draw conclusions about the functions involved, particularly concerning the behavior of the function F(x) derived from the integral expressions.

CECE2
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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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The OP has reformatted the original post.
 
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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