For x in R, let f(x)=|sin x| and g(x)=int_0^x | vedclass

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Question

(a)

Given, for $x \in R$

$f(x)=|\sin x|$

$g(x)=\int_0^x f(t) d t$, and

$p(x)=g(x)-\frac{2}{\pi} x$

$\because p(x+\pi)=g(x+\pi)-\frac{2}

Vedclass · Accepted Answer

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

Vedclass · Answer

(a)

Given, for $x \in R$

$f(x)=|\sin x|$

$g(x)=\int_0^x f(t) d t$, and

$p(x)=g(x)-\frac{2}{\pi} x$

$\because p(x+\pi)=g(x+\pi)-\frac{2}{\pi}(x+\pi)$

$=\int \limits_0^{x+\pi}|\sin (t)| d t-\frac{2}{\pi} x-2$

$=\int \limits_0^\pi|\sin t| d t+\int \limits_\pi^{\pi+x}|\sin t| d t-\frac{2}{\pi} x-2$

$=2 \int \limits_0^{\pi / 2} \sin t d t+\int \limits_0^x|\sin t| d t-\frac{2}{\pi} x-2$

$[\because|\sin t|$ is periodic function having period $\pi$ ]

$2[-\cos t]_0^{\pi / 2}+g(x)-\frac{2}{\pi} x-2$

$2(0-(-1))+g(x)-\frac{2}{\pi} x-2$

$2+g(x)-\frac{2}{\pi} x-2=g(x)-\frac{2}{\pi} x=p(x)$

$	herefore p(x+\pi)=p(x)$ for all $x$.

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

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