For x in mathbb{R},let f(x) = |sin x| and g(x | vedclass

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

Question

(A) Given $f(x) = |\sin x|$ and $g(x) = \int_0^x |\sin t| \, dt$.$p(x) = g(x) - \frac{2}{\pi} x$.We evaluate $p(x + \pi) = g(x + \pi) - \frac{2}{\pi}(

Vedclass · Accepted Answer

$p(x + \pi) = p(x)$ for all $x$

Vedclass · Answer

(A) Given $f(x) = |\sin x|$ and $g(x) = \int_0^x |\sin t| \, dt$.$p(x) = g(x) - \frac{2}{\pi} x$.We evaluate $p(x + \pi) = g(x + \pi) - \frac{2}{\pi}(x + \pi)$.$g(x + \pi) = \int_0^{x + \pi} |\sin t| \, dt = \int_0^{\pi} |\sin t| \, dt + \int_{\pi}^{x + \pi} |\sin t| \, dt$.Since $|\sin t|$ is periodic with period $\pi$,$\int_{\pi}^{x + \pi} |\sin t| \, dt = \int_0^x |\sin t| \, dt = g(x)$.Also,$\int_0^{\pi} |\sin t| \, dt = 2 \int_0^{\pi/2} \sin t \, dt = 2[-\cos t]_0^{\pi/2} = 2(0 - (-1)) = 2$.Thus,$g(x + \pi) = 2 + g(x)$.Substituting this into $p(x + \pi)$:$p(x + \pi) = (2 + g(x)) - \frac{2}{\pi} x - \frac{2}{\pi} \cdot \pi = 2 + g(x) - \frac{2}{\pi} x - 2 = g(x) - \frac{2}{\pi} x = p(x)$.Therefore,$p(x + \pi) = p(x)$ for all $x$.

For $x \in \mathbb{R}$,let $f(x) = |\sin x|$ and $g(x) = \int_0^x f(t) \, dt$. Let $p(x) = g(x) - \frac{2}{\pi} x$. Then:

Similar Questions

If $f(x) = f(a-x)$, then $\int_0^a x f(x) dx$ is equal to

Let $N$ be the set of natural numbers. For $n \in N$,define $I_n = \int_0^\pi \frac{x \sin^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx$. Then,for $m, n \in N$,which of the following is true?

The value of $\int_{0}^{\pi} \log (1+\cos x) d x$ is

$\int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$ is equal to

Let $I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$. Given that $\int \frac{d x}{1+k x^2}=\frac{1}{\sqrt{k}} \tan ^{-1}(\sqrt{k} x)+c, \tan ^{-1}(0)=0$ and $\tan ^{-1}(\sqrt{3})=\frac{\pi}{3}$. Then $3 I^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module