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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int_a^b x f(x) dx$. Using the property $\int_a^b g(x) dx = \int_a^b g(a + b - x) dx$,we get: $I = \int_a^b (a + b - x) f(a + b - x) dx$. Since it is given that $f(a + b - x) = f(x)$,we substitute this into the integral: $I = \int_a^b (a + b - x) f(x) dx$. $I = (a + b) \int_a^b f(x) dx - \int_a^b x f(x) dx$. $I = (a + b) \int_a^b f(x) dx - I$. $2I = (a + b) \int_a^b f(x) dx$. $I = \frac{a + b}{2} \int_a^b f(x) dx$.

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

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