If f(a+b-x)=f(x) then ∫ab xf(x)dx=

I=∫ab xf(x)dx→(i) Apply ∫abf(x) dx=∫abf(a+b-x) dx ⇒I=∫ab (a+b−x)f(a+b-x)dx I=∫ab(a+b-x)f(x)dx→(ii) (∵f(a+b-x)=f(x) ) (i)+(ii)⇒2I=∫abf(x)(x+a+b-x)dx 2I=(a+b)∫abf(x)dx ⇒I=(a+b)2∫abf(x)dx

If f(a+b-x)=f(x) then ∫ab xf(x)dx=

I=∫ab xf(x)dx→(i) Apply ∫abf(x) dx=∫abf(a+b-x) dx ⇒I=∫ab (a+b−x)f(a+b-x)dx I=∫ab(a+b-x)f(x)dx→(ii) (∵f(a+b-x)=f(x) ) (i)+(ii)⇒2I=∫abf(x)(x+a+b-x)dx 2I=(a+b)∫abf(x)dx ⇒I=(a+b)2∫abf(x)dx

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$I f f (a + b - x) = f (x)$ then $\int_{a}^{b} xf (x) dx =$

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

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

$\int_{a}^{b} f (x) dx$

answer is A.

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Detailed Solution

$\begin{aligned} I = \int_{a}^{b} xf (x) dx \to (i) \\ A p p l y \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \\ \Rightarrow I = \int_{a}^{b} (a + b - x) f (a + b - x) dx \\ I = \int_{a}^{b} (a + b - x) f (x) d x \to (i i) (∵ f (a + b - x) = f (x)) \\ (i) + (i i) \Rightarrow 2 I = \int_{a}^{b} f (x) (x + a + b - x) dx \\ 2 I = (a + b) \int_{a}^{b} f (x) dx \\ \Rightarrow I = \frac{(a + b)}{2} \int_{a}^{b} f (x) dx \end{aligned}$