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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.

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

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    $\begingroup$ Or the graph of $g(x)=f(a+b-x)$ is obtained from the graph of $f(x)$ by reflection in the line $x=\frac{a+b}{2}$. $\endgroup$ – N. S. Jul 4 '12 at 16:06
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Change of variables: $a+b-x=t$, $dx = -dt$, and $$ \int_a^b f(a+b-x)\, dx = -\int_b^a f(t)\, dt = \int_a^b f(t)\, dt. $$

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  • $\begingroup$ And why is the last one equal to $\int_a^b f(x)dx$? If $t=a+b-x$. $\endgroup$ – mavavilj Apr 19 '16 at 10:28
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    $\begingroup$ @mavavilj Dummy variables... $\endgroup$ – Siminore Apr 19 '16 at 12:00
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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*}

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  • $\begingroup$ But how is $\int_a^b f(x)dx = \int_a^b f(u)du$? $\endgroup$ – mavavilj Apr 19 '16 at 10:48
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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

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  • $\begingroup$ How is $dx=-d(a+b-x)$? $\endgroup$ – mavavilj Apr 19 '16 at 10:48
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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.

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

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

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