Question

If \[f\left( {a + b - x} \right) = f\left( x \right)\], then what is the value of \[\int\limits_a^b {xf\left( x \right)} dx\]?
A. \[\dfrac{{a + b}}{2}\int\limits_a^b {f\left( {b - x} \right)} dx\]
B. \[\dfrac{{a + b}}{2}\int\limits_a^b {f\left( x \right)} dx\]
C. \[\dfrac{{b - a}}{2}\int\limits_a^b {f\left( x \right)} dx\]
D. None of these

Answer 1

Hint: Here, a definite integral is given. First, apply the definite integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\] and simplify the integral. After that, substitute the given equation \[f\left( {a + b - x} \right) = f\left( x \right)\] and simplify the integral. Then, add this integral with the original integral and simplify it to get the required answer.

Formula Used: Definite integration rule: \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\]

Complete step by step solution: The given definite integral is \[\int\limits_a^b {xf\left( x \right)} dx\] and \[f\left( {a + b - x} \right) = f\left( x \right)\].
Let consider,
\[I = \int\limits_a^b {xf\left( x \right)} dx\] \[.....\left( 1 \right)\]
Now apply the definite integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\] on the right-hand side.
 \[I = \int\limits_a^b {\left( {a + b - x} \right)f\left( {a + b - x} \right)} dx\]
Substitute the given equation \[f\left( {a + b - x} \right) = f\left( x \right)\].
\[I = \int\limits_a^b {\left( {a + b - x} \right)f\left( x \right)} dx\]
\[ \Rightarrow I = \int\limits_a^b {\left( {a + b} \right)f\left( x \right)} dx - \int\limits_a^b {xf\left( x \right)} dx\] \[.....\left( 2 \right)\]
Add the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[2I = \int\limits_a^b {xf\left( x \right)} dx + \int\limits_a^b {\left( {a + b} \right)f\left( x \right)} dx - \int\limits_a^b {xf\left( x \right)} dx\]
\[ \Rightarrow 2I = \int\limits_a^b {\left( {a + b} \right)f\left( x \right)} dx\]
\[ \Rightarrow 2I = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)} dx\]
\[ \Rightarrow I = \dfrac{{a + b}}{2}\int\limits_a^b {f\left( x \right)} dx\]
Therefore,
\[\int\limits_a^b {xf\left( x \right)} dx = \dfrac{{a + b}}{2}\int\limits_a^b {f\left( x \right)} dx\]

Option ‘B’ is correct

Note: There are other definite integral rules that are derived from the rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx\].
The other rules are:
\[\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx\]
\[\int\limits_0^{2a} {f\left( x \right)} dx = \int\limits_0^a {f\left( x \right)} dx + \int\limits_0^{2a} {f\left( {2a - x} \right)} dx\]