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I've been assigned the following question:

Prove that if $g$ is integrable on $[a,b]$, then $\int_{a}^{b}g(x)dx=\int_{a}^{b}g(a+b-x)dx.$

I am stumped because it seems to me that

$$\int_{a}^{b}g(a+b-x)dx=-\int_{a}^{b}g(x)dx=\int_{b}^{a}g(x)dx.$$

Is the question wrong, or have I messed it all up?

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If you change variables to u = a + b - x, then the order of the limits changes, but also dx = -du. Hence the identity holds. –  Simon S Jan 19 '12 at 5:21
    
I am still unsure. Letting $u=a+b-x$ I get the following: $$\int_{a}^{b}g(x)dx=\int_{a}^{b}-g(u)du=-\int_{a}^{b}g(u)du =-(G(b)-G(a))=G(a)-G(b)$$ –  J.Borges Jan 19 '12 at 5:38
    
Geometrically, you are just reflecting the graph across the vertical line with $x$-coordinate $\frac{a+b}{2}$, so it is natural that the area under the curve doesn't change. –  Jonas Meyer Jan 19 '12 at 5:42
1  
The mistake you're making is not changing the limits. For u = a + b - x, when x = a, u = b; x = b, u = a. –  Simon S Jan 19 '12 at 5:43
1  
For heaven's sake, just do an example. Calculate $\int_2^3x\,dx$ and $\int_2^3(5-x)\,dx$ and tell me if you're still getting a minus sign in there. –  Gerry Myerson Jan 19 '12 at 8:50
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2 Answers

up vote 3 down vote accepted

Just put $z = a+b-x$, then you have $dz = - dx$, and when $x = a$ , $z = b$ and when $x=b, z =a$. So the integral becomes $$\int\limits_{b}^{a} g(z) \cdot - dz = \int\limits_{a}^{b} g(t) \ dt$$

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Here's a "picture proof":

Note that

  • The graph of $\color{maroon}{y=g(-x)}$ is the graph of $\color{darkgreen}{y=g(x)}$ reflected through the $y$-axis.

  • The graph of $\color{darkblue}{y=g\bigl(-(x-(a+b))\bigr)=g(a+b-x)}$ is the graph of $\color{maroon}{y=g(-x)}$ shifted to the right $a+b$ units.

enter image description here

Now think about the definite integral and it's relation to area.


Alternatively (this is just the substitution method):

If $G(x)$ is an antiderivative of $g(x)$, then $-G (a+b-x)$ is an antiderivative of $g(a+b-x)$; so $$ \int_a^b g(a+b-x)\,dx =-G(a+b-x)\bigl|_a^b =-\bigl(G(a)-G(b) \bigr) =G(b)-G(a)=\int_a^bg(x)\,dx. $$


But of course, since picture proofs aren't really proofs and since an integrable function does not necessarily have a primitive, the proper way to do this is to use Riemann sums. This would not be to difficult to do since any Riemann sum for $g(x)$ over $[a,b]$ corresponds to (by reversing the order of the summation) a Riemann sum for $g(a+b-x)$.

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