Prove that $ \int _a^b\:f\left(x\right)\:\:dx\:=\:\int _a^b\:f\left(a+b-x\right)\:\:dx $ This is the prove-
$y=a+b-x$. Then $dy/dx=-1$ and thus $$\int_a^bf(a+b-x)dx= \int_b^a-f(y)dy = \int_a^bf(x)dx$$
But I dont understand the last part of this -
Why ? $$\int_b^a-f(y)dy = \int_a^bf(x)dx$$
I know that $\int_b^a-f(y)dy = \int_a^bf(y)dy $. But something is not clear to me.
 A: Because the "$y$" and "$x$" are merely dummy variables. This is a definite integral, not an indefinite integral; the variables are only there to tell you that the function has one argument.
A definite integral is to be thought of as the "area under the graph", and so it doesn't matter what name the variable has, the area is still the same number.
$$\int_{a}^{b} f(x) \;dx = \int_{a}^{b}f(t) \;dt = \int_{a}^{b} f$$
In cases when the integration is implicitly understood (for example, if there is only one variable) it is not uncommon to see the variable(s) omitted altogether, as in the last case I wrote above.
A: This is simply because
$$
\int_a^b f(x)\; dx = \int_a^b f(y)\; dy = \int_a^b f(z)\; dz.
$$
A: You know that
$$\int_{b}^{a}-f(y)\text{ d}y = \int_{a}^{b}f(y)\text{ d}y\text{.}$$
But remember the one thing when it comes to integrals: the variable inside the integral is a dummy variable. That variable doesn't matter much at all, and regardless of what the variable is, you will get the same value. So for instance,
$$\int_{a}^{b}f(y)\text{ d}y = \int_{a}^{b}f(\text{anything})\text{ d}(\text{anything})\text{.}$$
It doesn't matter what $\text{anything}$ is.
