How is the Integral of $\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$ Can Some one tell me what this method is called and how it works With a detailed proof 
$$\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$$
I've been using this a lot in definite integration but haven't seemed to have realized why it is true.  But whatever it is it always seems to work.
Basically a proof of how it is always true.
 A: Let the antiderivative of $f$ be $F$.
Then $-\int_a^b f(a+b-x) d(a+b-x) = -(F(a+b-b) - F(a+b-a)) = F(b) - F(a) = \int_a^bf(x)dx$ .
EDIT Thank you for the correction avatar
A: Define $u=a+b-x$ so that $dx=-du$. Then the boundary term $x=a$ gives $u=b$ and $x=b$ gives $u=a$. Changing variables in the integral gives:
$$\int_a^bf(x) \, dx = -\int_b^af(u) \, du = \int_a^bf(u) \, du=\int_a^bf(x) \, dx$$
Intuititively, instead of integrating from $a$ to $b$, you are starting at $u=a+b-a=b$ and integrating left to $a$, but then switching sign to account for the fact that you were integrating leftwards.
A: Change of variables: $a+b-x=t$, $dx = -dt$, and
$$
\int_a^b f(a+b-x)\, dx = -\int_b^a f(t)\, dt = \int_a^b f(t)\, dt.
$$
A: Here is a pictorial argument.
$\displaystyle \int_a^b f(x) dx$ is the area under the curve $y=f(x)$ in the interval $(a,b)$ when you integrate from left to right.

$\displaystyle \int_a^b f(a+b-x) dx$ is the area under the curve $y=f(x)$ in the interval $(a,b)$ when you integrate from right to left.

Hence, both are equal.
A: it is just substitution, if we let $u = a+b-x$, we have $du = -dx$ and hence (note that $u = b$ when $x= a$ and vice versa)
\begin{align*}
  \int_a^b f(x)\,dx &= \int_a^b f(u)\, du\\
       &= \int_b^a f(a+b-x)\bigl(-dx\bigr)\\
       &= -\int_b^a f(a+b-x)\,dx\\
       &= \int_a^b f(a+b-x)\, dx
\end{align*}
A: Lets us define two functions
$G1(x) = F(x)$ and 
$G2(x) = F(a+b-x)$
For any point $x$ in the range $x=a$ to $x=b$
we can define a variable scalar $P$ such that $x$ divides the interval $a...b$ in the ratio $(P)$:$(1-P)$ where $0 <= P <=1$.
Now we can define any point $x$ in two ways:
$x1 = a + P(b-a)$ and 
$x2 = b - (1-P)(b-a)$ 
Now let us insert $x2$, $x1$ into the two different functions $G1$,$G2$
$G1(x) = F([x2])     = F([b - (1-P)(b-a)])     = F(a + P(b-a))$
$G2(x) = F(a+b-[x1]) = F(a + b - [a + P(b-a)]) = F(b - P(b-a))$
therefore
$\int_0^1G1(x)\,\mathrm{d}P =\int_0^1G2(x)\,\mathrm{d}P$
because in the former integration we move across the interval from $x=a$ to $x=b$ and in the latter integration we move across the same interval from $x=b$ to $x=a$.
Therefore
$\int_a^bF(x)$ = 
$\int_a^b F(a+b-x)$
A: Let $F(x)$ be the antiderivative of $f(x)$ with $\int f(x) \mathop{dx} = F(x)$. Using linear substitution:
$$\int f(g(x))\mathop{dx} = \frac{1}{g'(x)} \cdot F(g(x))$$
in which $g(x)$ is $a+b-x$.
You get:
$$
\int f(a+b-x) \mathop{dx} = \frac{1}{\frac{d}{dx} (a+b-x)} \cdot F(a+b-x) = -F(a+b-x)
$$
So if you now want to calculate $\int_a^b f(a+b-x) \mathop{dx}$ you get:
$$
\int_a^b f(a+b-x) \mathop{dx} = -F(a+b-b) +F(a+b-a) = F(b) - F(a) = \int_a^b f(x) \mathop{dx}
$$
