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

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

Explanation

Solution

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 uu in place of xx and then, consider u=a+bxu = a + b - x and differentiate it with respect to xx and put these values in place of uu to get the desired result.
Formulae to be used:
Change of variables: abf(x)dx=abf(u)du\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(u)du} ,
abf(x)dx=baf(x)dx- \int\limits_a^b {f(x)dx} = \int\limits_b^a {f(x)dx} .

Complete step by step answer:
Given the left-hand side abf(x)dx\int\limits_a^b {f(x)dx} .
To prove it is equal to abf(a+bx)dx\int\limits_a^b {f(a + b - x)dx} .
First, we can write abf(x)dx\int\limits_a^b {f(x)dx} as abf(u)du\int\limits_a^b {f(u)du} using the property of change of variables.
And then let u=a+bxu = a + b - x , and differentiate it with respect to xx , we get, dudx=1\dfrac{{du}}{{dx}} = - 1 i.e., du=dxdu = - dx .
And the limits will also change as we substitute the value of uu , when u=au = a , then, a=a+bxa = a + b - x i.e., x=bx = b and when u=bu = b , then, b=a+bxb = a + b - x i.e., x=ax = a .
Now, substituting the value of uu and the limits, we get, abf(u)du=baf(a+bx)(dx)\int\limits_a^b {f(u)du} = \int\limits_b^a {f(a + b - x)( - dx)} , which can also be written as abf(u)du=baf(a+bx)dx\int\limits_a^b {f(u)du} = - \int\limits_b^a {f(a + b - x)dx} .
Now, using the identity abf(x)dx=baf(x)dx- \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, baf(a+bx)dx=abf(a+bx)dx- \int\limits_b^a {f(a + b - x)dx} = \int\limits_a^b {f(a + b - x)dx} .
Hence, we proved that baf(a+bx)(dx)=abf(a+bx)dx\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 baf(a+bx)(dx)=abf(a+bx)dx\int\limits_b^a {f(a + b - x)( - dx)} = \int\limits_a^b {f(a + b - x)dx} is when limits are from 00 to aa , then the identity becomes 0af(x)dx=0af(ax)dx\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.