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If $\int \limits_0^{2 \pi} |x \: \sin x| dx=k\pi$, then the value of $k$ is equal to ______.
asked in Calculus | 2.4k 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.
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dv = sinxdx

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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, $\int_{0}^{2\pi}|xsinx|=\pi-(-3\pi)=4\pi$
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
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Awesome. Thank You.

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 $\pi +3\pi = 4\pi$

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.

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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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