Let f: mathbb{R} rightarrow mathbb{R} and g: | vedclass

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

Question

(D) Let $h(x) = [f(x)+f(-x)][g(x)-g(-x)]$. We evaluate $h(-x)$: $h(-x) = [f(-x)+f(x)][g(-x)-g(x)]$. This can be rewritten as: $h(-x) = [f(x)+f(-x)] \c

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $h(x) = [f(x)+f(-x)][g(x)-g(-x)]$. We evaluate $h(-x)$: $h(-x) = [f(-x)+f(x)][g(-x)-g(x)]$. This can be rewritten as: $h(-x) = [f(x)+f(-x)] \cdot -[g(x)-g(-x)]$. $h(-x) = -[f(x)+f(-x)][g(x)-g(-x)] = -h(x)$. Since $h(-x) = -h(x)$,the function $h(x)$ is an odd function. According to the property of definite integrals,if $h(x)$ is an odd function,then $\int_{-a}^{a} h(x) \, dx = 0$. Therefore,$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(x)+f(-x)][g(x)-g(-x)] \, dx = 0$.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: \mathbb{R} \rightarrow \mathbb{R}$ be continuous functions. Then the value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [f(x)+f(-x)][g(x)-g(-x)] \, dx$ is

Similar Questions

$\int_0^{\frac{\pi}{2}} \frac{300 \sin x+100 \cos x}{\sin x+\cos x} \,dx = \ldots$ (in $\pi$)

If $\int_{0}^{\pi} \log (\sin x) dx = 8 k$,then $\int_{0}^{\pi / 4} \log (1 + \tan x) dx =$

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

$\int_0^1 \frac{\log _e(1+x)}{1+x^2} d x=$

If $M = \int_{0}^{\pi / 2} \frac{\cos x}{x+2} dx$ and $N = \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{(x+1)^{2}} dx$, then the value of $M-N$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module