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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 | 3.2k views

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 (432k points)
edited
+1

dv = sinxdx

+2
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

$\int_{0}^{2\pi}|xsinx|=\int_{0}^{\pi}xsinx\,-\int_{\pi}^{2\pi}xsinx$

$[sinx-xcosx]_{0}^{\pi}=\pi$

and

$[sinx-xcosx]_{\pi}^{2\pi}=-3\pi$

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

I am getting pi only
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 $I=\int_{0}^{\pi }(\pi-x)sinxdx$

$2I=\pi\int_{0}^{\pi}sinxdx=\pi$

similarly from 2nd part we can get $3\pi$ so total $4\pi$ is the answer
by Active (4.1k points)
0
0
Awesome. Thank You.
+4

$f(x)=sinx$

$f(x)=cosx$

ILATE PROPERTY IS USEFULL TO DO SUCH TYPE OF QUESTION

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)

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 (59.5k points)
edited

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.

by Active (1.2k points)
0
We can use the property of definite integral and apply the formula for limits and solve instead of byparts method too