If f(a+b−x)=f(x), then ∫abxf(x)dx=

a + b 2 ∫ b a f ( a + b − x ) d x

a + b 2 ∫ b a f ( a + b − x ) d x

a + b 2 ∫ a b f ( b − x ) d x

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If $f (a + b - x) = f (x),$ then $\int_{a}^{b} x f (x) d x =$

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$\frac{a + b}{2} \int_{b}^{a} f (a + b - x) d x$

$\frac{b - a}{2} \int_{a}^{b} f (x) d x$

$\frac{a + b}{2} \int_{a}^{b} f (x) d x$

$\frac{a + b}{2} \int_{a}^{b} f (b - x) d x$

answer is A.

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CBSE

$f (a + b - x) = f (x);$ removable of $x$

$I = \int_{a}^{b} f (x) d x$

$I = \int_{a}^{b} (a + b - x) f (a + b - x) d x = \frac{a + b}{2} \int_{a}^{b} f (x) d x$

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