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Question
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
Solution
\[\int_a^b xf\left( x \right)dx\]
\[ = \int_a^b \left( a + b - x \right)f\left( a + b - x \right)dx ..................\left[ \int_a^b f\left( x \right)dx = \int_a^b f\left( a + b - x \right)dx \right]\]
\[ = \int_a^b \left( a + b - x \right)f\left( x \right)dx ..................\left[ f\left( a + b - x \right) = f\left( x \right) \right]\]
\[ \therefore \int_a^b xf\left( x \right)dx = \int_a^b \left( a + b \right)f\left( x \right)dx - \int_a^b xf\left( x \right)dx\]
\[\Rightarrow \int_a^b xf\left( x \right)dx + \int_a^b xf\left( x \right)dx = \left( a + b \right) \int_a^b f\left( x \right)dx\]
\[ \Rightarrow 2 \int_a^b xf\left( x \right)dx = \left( a + b \right) \int_a^b f\left( x \right)dx\]
\[ \Rightarrow \int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
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