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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.

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.

Change of variables: $a+b-x=t$, $dx = -dt$, and
$$
int_a^b f(a+b-x), dx = -int_b^a f(t), dt = int_a^b f(t), dt.
$$

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(-dxbigr)\
&= -int_b^a f(a+b-x),dx\
&= int_a^b f(a+b-x), dx
end{align*}

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

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.

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
$int_a^bF(x)$ =
$int_a^b F(a+b-x)$

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:
$$
int_a^b f(a+b-x) mathop{dx} = -F(a+b-b) +F(a+b-a) = F(b) - F(a) = int_a^b f(x) mathop{dx}
$$