Let f:R to R and g:R to R be continuous funct | vedclass

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

Question

(D) Let $h(x) = [f(x) + f(-x)][g(x) - g(-x)]$. Now,calculate $h(-x)$: $h(-x) = [f(-x) + f(-(-x))][g(-x) - g(-(-x))] = [f(-x) + f(x)][g(-x) - g(x)]$. W

Vedclass · Accepted Answer

$ 0 $

Vedclass · Answer

(D) Let $h(x) = [f(x) + f(-x)][g(x) - g(-x)]$. Now,calculate $h(-x)$: $h(-x) = [f(-x) + f(-(-x))][g(-x) - g(-(-x))] = [f(-x) + f(x)][g(-x) - g(x)]$. We can rewrite this as: $h(-x) = -[f(x) + f(-x)][g(x) - g(-x)] = -h(x)$. Since $h(-x) = -h(x)$,the function $h(x)$ is an odd function. For any odd function $h(x)$,the integral over a symmetric interval $[-a, a]$ is zero: $\int_{-a}^{a} h(x) \, dx = 0$. Therefore,$\int_{-\pi/2}^{\pi/2} [f(x) + f(-x)][g(x) - g(-x)] \, dx = 0$.

Let $f:R \to R$ and $g:R \to R$ be continuous functions,then the value of the integral $\int_{-\pi/2}^{\pi/2} [f(x) + f(-x)][g(x) - g(-x)] \, dx$ is:

Similar Questions

For $I_n = \int_{1}^{e} (\ln x)^n dx$,where $n \in N$,which of the following relations holds true?

The value of $\int_{0}^{1} \frac{dx}{x + \sqrt{1 - x^2}}$ is

Let $[\cdot]$ denote the greatest integer function. Then $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{12(3+[x])}{3+[\sin x]+[\cos x]} \right) dx$ is equal to:

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$ is equal to . . . . . . .

$\int_{\pi / 11}^{9 \pi / 22} \frac{d x}{1+\sqrt{\tan x}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module