Question

Prove that $\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(a + b - x)dx} $ .

Answer 1

Hint: In this question, we have to prove the left- hand side is equal to the right- hand side. For this, we will consider the left- hand side and simplify it to prove it is equal to the right- hand side.
It is a common identity which we use to solve any question and here we have to prove this identity.
So first, we will use the identity to change variables and substitute $u$ in place of $x$ and then, consider $u = a + b - x$ and differentiate it with respect to $x$ and put these values in place of $u$ to get the desired result.
Formulae to be used:
Change of variables: $\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(u)du} $ ,
$ - \int\limits_a^b {f(x)dx} = \int\limits_b^a {f(x)dx} $ .

Complete step by step answer:
Given the left-hand side $\int\limits_a^b {f(x)dx} $ .
To prove it is equal to $\int\limits_a^b {f(a + b - x)dx} $ .
First, we can write $\int\limits_a^b {f(x)dx} $ as $\int\limits_a^b {f(u)du} $ using the property of change of variables.
And then let $u = a + b - x$ , and differentiate it with respect to $x$ , we get, $\dfrac{{du}}{{dx}} = - 1$ i.e., $du = - dx$ .
And the limits will also change as we substitute the value of $u$ , when $u = a$ , then, $a = a + b - x$ i.e., $x = b$ and when $u = b$ , then, $b = a + b - x$ i.e., $x = a$ .
Now, substituting the value of $u$ and the limits, we get, $\int\limits_a^b {f(u)du} = \int\limits_b^a {f(a + b - x)( - dx)} $ , which can also be written as $\int\limits_a^b {f(u)du} = - \int\limits_b^a {f(a + b - x)dx} $ .
Now, using the identity $ - \int\limits_a^b {f(x)dx} = \int\limits_b^a {f(x)dx} $ , we can reverse the limits which will remove the negative sign, we get, $ - \int\limits_b^a {f(a + b - x)dx} = \int\limits_a^b {f(a + b - x)dx} $ .
Hence, we proved that $\int\limits_b^a {f(a + b - x)( - dx)} = \int\limits_a^b {f(a + b - x)dx} $

Note:
One must know the basic identities associated with integration while solving such questions or proving such identities.
One special case using the identity $\int\limits_b^a {f(a + b - x)( - dx)} = \int\limits_a^b {f(a + b - x)dx} $ is when limits are from $0$ to $a$ , then the identity becomes $\int\limits_0^a {f(x)dx} = \int\limits_0^a {f(a - x)dx} $ .
Remember to change the limits, when you substitute any variable and differentiate it with respect to the variable.