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

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

Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int_{a}^{b} x \cdot f(a + b - x) \, dx$. Using the property $\int_{a}^{b} g(x) \, dx = \int_{a}^{b} g(a + b - x) \, dx$,we get: $I = \int_{a}^{b} (a + b - x) \cdot f(a + b - (a + b - x)) \, dx$ $I = \int_{a}^{b} (a + b - x) \cdot f(x) \, dx$ $I = (a + b) \int_{a}^{b} f(x) \, dx - \int_{a}^{b} x \cdot f(x) \, dx$. Since $f(a + b - x) = f(x)$,the original integral is $I = \int_{a}^{b} x \cdot f(x) \, dx$. Substituting this into the equation: $I = (a + b) \int_{a}^{b} f(x) \, dx - I$ $2I = (a + b) \int_{a}^{b} f(x) \, dx$ $I = \frac{a + b}{2} \int_{a}^{b} f(x) \, dx$.

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

Similar Questions

If $P = \int_0^{3\pi} f(\cos^2 x) dx$ and $Q = \int_0^{\pi} f(\cos^2 x) dx$,then:

Assertion $(A)$: $\int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} [2 \sin x] dx = 0$,where $[.]$ denotes the greatest integer function.
Reason $(R)$: $2 \sin x$ is a decreasing function in $\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$.

The value of the integral $\int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{-101}} d x$ is equal to

If $\int_0^\pi {xf(\sin x)dx = A} \int_0^{\pi /2} {f(\sin x)dx} $,then $A$ is

The value of the integral $\int \limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x$ is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module