If f(a+b-x)=f(x),then int_a^b x f(x) d x= . | vedclass

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(A) Let $I = \int_a^b x f(x) d x$. Using the property $\int_a^b g(x) d x = \int_a^b g(a+b-x) d x$,we get: $I = \int_a^b (a+b-x) f(a+b-x) d x$. Since $

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int_a^b x f(x) d x$. Using the property $\int_a^b g(x) d x = \int_a^b g(a+b-x) d x$,we get: $I = \int_a^b (a+b-x) f(a+b-x) d x$. Since $f(a+b-x) = f(x)$,this becomes: $I = \int_a^b (a+b-x) f(x) d x$. $I = (a+b) \int_a^b f(x) d x - \int_a^b x f(x) d x$. $I = (a+b) \int_a^b f(x) d x - I$. $2I = (a+b) \int_a^b f(x) d x$. $I = \frac{a+b}{2} \int_a^b f(x) d x$. Thus,the correct option is $A$.

If $f(a+b-x)=f(x)$,then $\int_a^b x f(x) d x=$ . . . . . . .

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