Integral Question (with a possible error) 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?
 A: 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 $$\int\limits_{b}^{a} g(z) \cdot - dz = \int\limits_{a}^{b} g(t) \ dt$$
A: 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.

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 $g(a+b-x)$.
