If `f(a+b-x)=f(x),\ t h e n\ int_a^b xf(x)dx` is equal to `(a+b)/2int_a^bf(b-x)dx` b. `(a+b)/2int_a^bf(b+x)dx` c. `(b-1)/2int_a^bf(x)dx` d. `(a+b)/2int_a^bf(x)dx`

Explanation

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

`text { Let } I=int_{a}^{b} x f(x) d x ldots ldots ldots ldots .(1) `

`=int_{a}^{b}(a+b-x) f(a+b-x) d x `

`=int_{a}^{b}(a+b-x) f(x) d x ldots ldots ldots ldots ldots .(2)`


Adding (1) and (2)

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

`=(a+b) int_{a}^{b} f(x) d x`

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