If f(a + b - x) = f(x), then int_{a}^{b} x f( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I = \int_{a}^{b} x f(x) dx$  --- $(1)$Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,we have:$I = \int_{a}^{b} (a+b-x)

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int_{a}^{b} x f(x) dx$  --- $(1)$Using the property $\int_{a}^{b} g(x) dx = \int_{a}^{b} g(a+b-x) dx$,we have:$I = \int_{a}^{b} (a+b-x) f(a+b-x) dx$Since $f(a+b-x) = f(x)$,the integral becomes:$I = \int_{a}^{b} (a+b-x) f(x) dx$Expanding the integral:$I = (a+b) \int_{a}^{b} f(x) dx - \int_{a}^{b} x f(x) dx$Substituting $I$ from equation $(1)$:$I = (a+b) \int_{a}^{b} f(x) dx - I$Adding $I$ to both sides:$2I = (a+b) \int_{a}^{b} f(x) dx$Dividing by $2$:$I = \frac{a+b}{2} \int_{a}^{b} f(x) dx$Thus,the correct option is $C$.

If $f(a + b - x) = f(x),$ then $\int_{a}^{b} x f(x) dx$ is equal to

Similar Questions

For any integer $n$, the value of the integral $\int_0^\pi e^{\cos^2 x} \cos^3(2n+1)x \, dx$ is:

Value of the definite integral $\int_{-1/2}^{1/2} (\sin^{-1}(3x - 4x^3) - \cos^{-1}(4x^3 - 3x)) dx$

$\int_{-1}^1 \left(\sqrt{1+x+x^2}-\sqrt{1-x+x^2}\right) dx =$

$\int_2^4 \frac{\log x^2}{\log x^2+\log (36-12x+x^2)} dx$ is equal to

If $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$,then $[I] = \ldots$. Here,$[.]$ denotes the greatest integer function.

Vedclass Test Series

Exam Paper Generator

Online Exam Module