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If $\int \limits_0^{2 \pi} |x \: \sin x| dx=k\pi$, then the value of $k$ is equal to ______.
in Calculus by Veteran (105k points) | 3.2k views

5 Answers

+40 votes
Best answer
There is a mod term in the given integral. So, first we have to remove that. We know that $x$ is always positive here and $\sin x$ is positive from 0 to $\pi$. From $\pi$ to $2\pi$, $x$ is positive while $\sin x$ changes sign. So, we can write

$\int_{0}^{2\pi} \mid x \sin x \mid dx = \int_{0}^{\pi} x \sin x dx + \left(-\int_{\pi}^{2\pi} x \sin x dx \right)$

$\qquad\qquad\qquad\quad\;= \int_{0}^{\pi} x \sin x dx -\int_{\pi}^{2\pi} x \sin x dx.$

$\int_{0}^{\pi}{udv} = uv - \int_{0}^{\pi}{vdu}$

$ \text{Here }u = x, du = dx, dv = \sin x dx, \text{ so } v = -\cos x$

$ \therefore \int_{0}^{\pi} x \sin x dx = \left [ -x\cos x\right]_0^{\pi} + \int_{0}^{\pi} \cos x dx$

$= \pi + \left[\sin x\right]_0^{\pi}\\ =\pi$

$\text{Now, } \int_{\pi}^{2\pi} x \sin x = \left [ -x\cos x\right]_{\pi}^{2\pi} + \int_{\pi}^{2\pi} \cos x dx$

$= -3\pi + \left[\sin x\right]_{\pi}^{2\pi}\\ = -3 \pi$

So, given integral = $\pi - \left(-3\pi\right) = 4\pi$

So, k = 4.
by Veteran (431k points)
edited by

dv = sinxdx

First, we need to defined function $F(x)=|xsin(x)|$

$f(x)=\begin{cases} xsinx & \text{ if } 0 \leq x \leq \pi \\ -xsinx & \text{ if } \pi \leq x \leq 2\pi \end{cases}$

Now, first only find $\int xsinx\,dx=sinx-xcosx$

Now, our integral becomes





So, $\int_{0}^{2\pi}|xsinx|=\pi-(-3\pi)=4\pi$
Can you explain how could you determinal -3pi

I am getting pi only
+9 votes
No need to apply integration by parts,it will be time consuming,

Appy the formula $\int_{0}^{\pi }(\pi-x)sinxdx+\int_{\pi}^{2\pi}(2\pi+\pi -x)sinxdx$

now $I=\int_{0}^{\pi }(\pi-x)sinxdx$


similarly from 2nd part we can get $3\pi$ so total $4\pi$ is the answer
by Active (4.1k points)
please elaborate ur method
Awesome. Thank You.




+4 votes


BUT ONE ANOTHER method is also important and fast when function is given odd function because both linear algebric and sin function is odd so their product is also odd function and period of sin is \pi and mod function converts them into positive so 

\int_{0}^{\pi }xsinx   gives value  \pi   and   \int_{\pi }^{2\pi } xsinxdx  gives value -3\pi but due to mod function it is gives\left | -3\pi \right |  that is equal to 3\pi    so final answer will be \pi +3\pi = 4\pi

by Active (5.1k points)
+2 votes

Some important things, we should know

  • ${\color{Red} {\cos 0 = 1,\cos \pi = -1,\cos 2\pi = 1,\dots}}$
  • ${\color{Blue}{ \text{In general}\: \cos n\pi = (-1)^{n}\: \text{where}\: n=0,1,2,\dots}}$
  • ${\color{Magenta} {\sin 0 = 0,\sin \pi = 0,\sin 2\pi = 0,\dots}}$
  • ${\color{Green} {\text{In general}\: \sin n\pi = 0\: \text{where}\: n=0,1,2,\dots}}$
  • ${\color{Orange} {\sin(\pi-x)=\sin x,\sin (2\pi-x)=-\sin x,\sin(3\pi-x)=\sin x}}$
  • ${\color{Teal} {\text{In general}\: \sin (n\pi-x)=(-1)^{n+1}\sin x\: \text{ where} \: n=0,1,2,\dots}}$
  •  ${\color{Orchid} {\cos(\pi-x)=-\cos x,\cos (2\pi-x)=\cos x,\cos (3\pi-x)=-\cos x}}$
  • ${\color{purple} {\text{ In general}\: \cos (n\pi-x)=(-1)^{n}\cos x \: \text{where}\: n=0,1,2,\dots}}$                                             

$f(x)=\sin x$

$f(x)=\cos x$


Let $I = \displaystyle{}\int_{0}^{2\pi} \mid x \sin x \mid {\mathrm{d} x} = k\pi\rightarrow(1)$

Now, we should break the limits and open the 'modulus function'.

$I = \displaystyle{}\int_{0}^{\pi} x \sin x \: {\mathrm{d} x} - \displaystyle{}\int_{\pi}^{2\pi} x \sin x\:  {\mathrm{d} x}$

Now, $I = \left[-x \cos x + \sin x \right]_{0}^{\pi} - \left[-x \cos x + \sin x \right]_{\pi}^{2\pi}$

$\implies I = \left[ \left(- \pi \cos \pi + \sin \pi \right) - \left(-0 \cos 0 + \sin 0 \right) \right] - \left[ \left(- 2\pi \cos 2\pi + \sin 2\pi \right) - \left(- \pi \cos \pi + \sin \pi \right) \right]$

$\implies I = \left[ \left(- \pi (-1) + 0 \right) - \left( 0 + 0 \right) \right] - \left[ \left(- 2\pi (1)   + 0  \right) - \left(- \pi (-1) + 0 \right) \right]$

$\implies I = \pi  + 2\pi + \pi $

$\implies I = 4\pi = k\pi\implies k = 4$

So, the correct answer is $k  = 4.$

by Veteran (59.2k points)
edited by
0 votes

Split in two intervals: 0 to π and π to 2kπ as sin x is negative in π to 2π.

Then use Integration By Parts with ILATE sequence.

Answer is:   4

by Active (1.2k points)
We can use the property of definite integral and apply the formula for limits and solve instead of byparts method too
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