# Thread: Proving a Theorem on Integrals

1. ## Proving a Theorem on Integrals

I'm clueless when it comes to proving theorems. It's frustrating that I don't really know what I'm expected to do. I can draw diagrams, and test the theorem for specific functions, but I don't know where to begin on proving the theorem in general.
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Theorem

If $\displaystyle f$ is continuous on $\displaystyle \Re$, prove that

$\displaystyle \int_{a}^{b}f(-x)dx=\int_{-b}^{-a}f(x)dx$

for the case that $\displaystyle f(x)\geq 0$ and $\displaystyle 0<a<b$.

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I'm clueless when it comes to proving theorems. It's frustrating that I don't really know what I'm expected to do. I can draw diagrams, and test the theorem for specific functions, but I don't know where to begin on proving the theorem in general.
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Theorem

If $\displaystyle f$ is continuous on $\displaystyle \Re$, prove that

$\displaystyle \int_{a}^{b}f(-x)dx=\int_{-b}^{-a}f(x)dx$

for the case that $\displaystyle f(x)\geq 0$ and $\displaystyle 0<a<b$.

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Just make the substitution u = -x... Of course, you can prove using Riemann Sums as well, but it's longer.

Tonio

3. Originally Posted by tonio
Just make the substitution u = -x... Of course, you can prove using Riemann Sums as well, but it's longer.

Tonio
Ok, I have $\displaystyle \int_{a}^{b}f(-x)dx=-\int_{a}^{b}f(u)du=\int_{b}^{a}f(u)du$

The only thing I can think of next, is to claim that $\displaystyle \int_{b}^{a}f(u)du=\int_{-b}^{-a}f(-u)du$ which is just $\displaystyle \int_{-b}^{-a}f(x)dx$ since $\displaystyle u=-x$.

Is this valid?

Ok, I have $\displaystyle \int_{a}^{b}f(-x)dx=-\int_{a}^{b}f(u)du=\int_{b}^{a}f(u)du$
The only thing I can think of next, is to claim that $\displaystyle \int_{b}^{a}f(u)du=\int_{-b}^{-a}f(-u)du$ which is just $\displaystyle \int_{-b}^{-a}f(x)dx$ since $\displaystyle u=-x$.