How is the Integral of $\int_a^bf(x)~dx~~=\int_a^bf(a+b-x)~dx$

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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*}

Answer 4

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.

Answer 5

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