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If F ( a + B − X ) = F ( X ) , Then Prove that ∫ B a X F ( X ) D X = ( a + B 2 ) ∫ B a F ( X ) D X

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Question

If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that

\[\int_a^b xf\left( x \right)dx = \left( \frac{a + b}{2} \right) \int_a^b f\left( x \right)dx\]
Sum
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Solution

\[\int_a^b xf\left( x \right)dx = \int_a^b \left( a + b - x \right)f\left( a + b - x \right)dx ................\left[ \int_a^b f\left( x \right)dx = \int_a^b f\left( a + b - x \right)dx \right]\]
\[ \Rightarrow \int_a^b xf\left( x \right)dx = \int_a^b \left( a + b - x \right)f\left( x \right)dx .....................\left[ f\left( a + b - x \right) = f\left( x \right) \right]\]
\[ \Rightarrow \int_a^b xf\left( x \right)dx = \int_a^b \left( a + b \right)f\left( x \right)dx - \int_a^b xf\left( x \right)dx\]
\[ \Rightarrow 2 \int_a^b xf\left( x \right)dx = \left( a + b \right) \int_a^b f\left( x \right)dx\]
\[ \Rightarrow \int_a^b xf\left( x \right)dx = \left( \frac{a + b}{2} \right) \int_a^b f\left( x \right)dx\]

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Chapter 19: Definite Integrals - Exercise 20.5 [Page 96]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.5 | Q 47 | Page 96

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