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In this question, people are saying that the definite integral of $f(x)$ from $0$ to $a$ is equal to the integral of $f(a-x)$ from $0$ to $a$. How can that be true? Simple examples don't work.

Community
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Ovi
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    What simple example do you have in mind? – ronno May 25 '13 at 19:23
  • If you post a "simple example" which doesn't work, then we can probably see exactly where you're going wrong, and give a much more helpful answer! – not all wrong May 25 '13 at 19:28
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    The area below a curve is the same if you reflect the curve in a vertical line... – Per Alexandersson May 25 '13 at 19:29
  • Well take for example f(x)=x. The integral from 0 to 5 is 25/2. Now take f(x-2.5)=x-2.5. The integral of this from 0 to 5 is equal to 0. – Ovi May 25 '13 at 19:31
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    Well, you're not following your own rules. You have to look at $f(5-x)$ !! – Ted Shifrin May 25 '13 at 19:32
  • Oh wow haha I must've been dreaming when I wrote this question, I completely didn't take into account that the a in a-x is the same as the a in the limit of integration, I thought a was just any arbitrary constant. – Ovi May 25 '13 at 23:56

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Put $z=a-x$ then $dz=-dx$. When $x=0, z=a$ and when $x=a,z=0$. Now $$\int_0^af(a-x)dx=\int_a^0f(z)(-dz)=\int_0^af(z)dz$$

pritam
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Consider the function $$g(x):=f(a-x)\qquad(0\leq x\leq a)\ .$$ It follows that $$g\left({a\over2}+y\right)=f\left({a\over2}-y\right) \qquad\left(-{a\over2}\leq y\leq{a\over2}\right)\ .$$ Therefore the graph of $g$ is the graph of $f$, reflected at the vertical $x={a\over2}$. Assuming for simplicity that $f$, and therefore $g$, are $\geq0$, it is obvious that the areas $A_f$ and $A_g$ in question are the same.

Christian Blatter
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