How can I prove that $\int_0^af(x) \; dx=\int_0^a f(a-x) \; dx$
How to prove $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$
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$\begingroup$
calculus
integration
3 Answers
7
It's pedagogically better to give hints:
Change of variables: If $a-x = u$ then $f(a-x) = f(u)$
Change of variables: If $a-x = u$ then $dx = -du$
Change of variables: If $a-x = u$ then $x =0 \iff u =\ \color{red}{??}$ and $x=a \iff u =\ \color{red}{??}$
Interchange of boundary: $ \displaystyle\int_a^{b} f(x) dx = - \displaystyle\int_{b}^{a} f(x) dx$
Formal variable renaming: $ \displaystyle\int_a^{b} f(x) dx = \displaystyle\int_{a}^{b} f(z) dz$
Can you fill in the $\color{red}{??}$ and put it all together?
3
Use change of variables $u=a-x$
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0@Rawhi: I was surprised that you asked 'How this could help!!?', that's all. – 2012-03-17
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When you think about the function $f(a-x)$ as a reflection of the 'original region' about the line $x=\frac{a}{2}$, then the result becomes rather apparent...