Question

$\mathfrak\green{Question:-}$

$\longmapsto\rm{Evaluate \: \displaystyle \int ^{2\pi} _{0} \rm | \sin(x) | \: dx}$



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Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle \int ^{2\pi} _{0} \rm | sinx | \: dx$

can be rewritten as

$\rm \: = \: \displaystyle \int ^{\pi} _{0} \rm | sinx | \: dx + \displaystyle \int ^{2\pi} _{\pi} \rm | sinx | \: dx$

$\bigg[\rm \because \: \displaystyle \int ^{b} _{a} f(x) dx = \displaystyle \int ^{c} _{a} f(x)dx + \displaystyle \int ^{b} _{c} f(x) dx \: \: (a \leqslant c \leqslant b )\: \bigg]$

We know,

By definition of Modulus function

$\begin{gathered}\begin{gathered}\bf\: |x| = \begin{cases} &\sf{ \: \: x \: \: if \: \: x \geqslant 0} \\ \\ &\sf{ - x \: if \: x \: < \: 0} \end{cases}\end{gathered}\end{gathered}$

So, by using this definition

$\begin{gathered}\begin{gathered}\bf\: |sinx| = \begin{cases} &\sf{ \: \: sinx \: \: if \: \: 0 \leqslant x \leqslant \pi} \\ \\ &\sf{ - sinx \: if \: \pi \leqslant x \leqslant 2\pi} \end{cases}\end{gathered}\end{gathered}$

So, using this result, the above expression can be rewritten as

$\rm \: = \: \displaystyle \int ^{\pi} _{0} \rm sinx \: dx - \displaystyle \int ^{2\pi} _{\pi} \rm sinx \: dx$

We know,

$\boxed{\sf{  \:\displaystyle \int \rm \: \: sinx \: dx \: = \: - \: cosx \: + \: c \: }}$

So, using this result, we get

$\rm \: =  \:\bigg( - cosx\bigg)^{\pi} _{0} - \bigg( -cosx \bigg)^{2\pi} _{\pi}$

$\rm \: =  \: - \: \bigg(cosx\bigg)^{\pi} _{0} + \bigg(cosx \bigg)^{2\pi} _{\pi}$

$\rm \: =  \: - (cos\pi - cos0) + (cos2\pi - cos\pi)$

We know,

$\boxed{\sf{  \: \: cosn\pi \: = \: {( - 1)}^{n} \: \: }}$

So, using this, we get

$\rm \: =  \: - ( - 1 - 1) + (1 + 1)$

$\rm \: =  \: 2 + 2$

$\rm \: =  \: 4$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: \displaystyle \int ^{2\pi} _{0} \rm | \sin(x) | \: dx \: = \: 4 \: \: }}$

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Additional Information:-

$\boxed{\sf{  \:\displaystyle \int ^{a} _{0} f(x)dx = \displaystyle \int ^{a} _{0} f(a - x) \: dx \: }}$

$\boxed{\sf{  \:\displaystyle \int ^{b} _{a} f(x)dx = \displaystyle \int ^{b} _{a} f(a + b- x) \: dx \: }}$

$\boxed{\sf{  \:\displaystyle \int ^{b} _{a} f(x)dx = \displaystyle \int ^{b} _{a} f(y) \: dy \: }}$

$\boxed{\sf{  \:\displaystyle \int ^{b} _{a} f(x)dx = - \: \displaystyle \int ^{a} _{b} f(x) \: dx \: }}$

$\boxed{\sf{  \:\displaystyle \int ^{a} _{ - a} f(x)dx = \: 2\displaystyle \int ^{a} _{0} f(x) \: dx \:\rm \: if \: f( - x) = f(x) }}$

$\boxed{\sf{  \:\displaystyle \int ^{ a} _{ - a} f(x)dx = \: 0 \:\rm \: if \: f( - x) = - f(x) }}$

$\boxed{\sf{  \:\displaystyle \int ^{2a} _{ 0} f(x)dx = \: 2\displaystyle \int ^{a} _{0} f(x) \: dx \:\rm \: if \: f( 2a- x) = f(x) }}$

$\boxed{\sf{  \:\displaystyle \int ^{ 2a} _{0} f(x)dx = \: 0 \:\rm \: if \: f(2a - x) = - f(x) }}$

Answer 2

$\color{green} \begin{array} {l}\text{We know that \( \sin x \) is positive, when \( 0 \leq x \leq \pi \)} \\ \text{ and \( \sin x \) is negative when \( \pi \leq x \leq 2 \pi \).} \end{array}$

$\red{ \tt\therefore \quad|\sin x|=\left\{\begin{aligned} \tt \sin x, & \text { when } \tt 0 \leq x \leq \pi \\ \tt-\sin x, & \text { when } \tt \pi \leq x \leq 2 \pi . \end{aligned}\right.}$

$\color{darkcyan}\begin{aligned} \tt \therefore \int_{0}^{2 \pi}|\sin x| d x & \tt=\int_{0}^{\pi}|\sin x| d x+\int_{\pi}^{2 \pi}|\sin x| d x \\ \\ & \tt=\int_{0}^{\pi} \sin x d x+\int_{\pi}^{2 \pi}(-\sin x) d x \\ \\ & \tt=\bigg[-\left.\cos x\bigg]_{0} ^{\pi}+ \bigg[\cos x \bigg]_{\pi}^{2 x}=(2+2)=4 \right.\end{aligned}$