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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate: $ \int \limits_{-1}^{3/2} \; \left | x \sin \pi x \right | \; dx $

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Toolbox:
  • $\; \int \limits_a^b f(x)dx=F(b)-F(a)$
Step 1:
Given $I= \int \limits_{-1}^{3/2} |x \sin \pi x|dx$
Let $\pi x=t$ on differentiating w.r.t x
$\pi.dx=dt$ Also $x=\pi/t,$ and $dx=dt/\pi$
The limit changes when we substitute t
When $x=-1,t=-\pi$ and
When $4x=3/2,\; t=3 \pi/2$
Therefore $I= \large\frac{1}{\pi^2} $$\int \limits_{-\pi}^{3\pi/2} |\large\frac{t}{\pi}$$ \sin t|\large\frac{dt}{\pi}$
$= \large\frac{1}{\pi^2} $$\int \limits_{-\pi}^{3\pi/2} |t \sin t|dt$
$= \large\frac{1}{\pi^2} $$\int \limits_{-\pi}^{\pi} | t\sin t|dt+ \int \limits_{\pi}^{3 \pi/2} |t\sin t|dt$$
$\sin t$ changes in the 3 rd and 4 th quadrant.
Now let us consider $f(t)=|t \sin t|$ on $\pi,3\pi/2$ (ie) in the 3 rd quadrant
Hence $ t \sin t=-ve$ therefore $|t \sin t|=-t \sin t$ on $[\pi,3\pi/2]$
Also Let $f(t) =|t \sin t|$
Therefore $ f(-t)=|-t. \sin (-t)|= t.\sin t$
Hence it is an even function
Therefore $ \int \limits_{-\pi}^{\pi} |t \sin t|dt= 2 \int \limits_0^{\pi} | t \sin t|dt =2 \int \limits_0^{\pi} t \sin t .dt$
Therefore $ I=\bigg\{ \frac{1}{\pi^2} \bigg( 2\int \limits_0^{\pi} t \sin t dt \bigg)- \int \limits_\pi^{3 \pi/2} t \sin t dt\bigg\}$
Step 2:
Consider $\int t \sin t\;dt$
Clearly this is of the form $\int udv,$ which can be solved by the method of integration
$\int udv=uv-\int vdu$
Let $u=t$ on differentiating w.r.t t,$du=dt$
Let $dv=\sin t dt $ on integrating we get, $v=-\cos t dt$
Therefore $ I= (-t \cos t)-\int -\cos t dt$
$=-(t \cos t)+\int \cos t dt$
On integrating we get
$-( t\cos t)+(\sin t)$
Step 3:
Therefore $ I=\large \frac{1}{\pi^2}$$ \bigg[2 \int \limits_0^{\pi} t \sin t dt -\int \limits_{\pi}^{3\pi/2} t \sin tdt\bigg]$
$= \large\frac{1}{\pi^2} $$\bigg[2( \sin t-t \cos t)\bigg]_0^{\pi}- \bigg[ \sin t- t cos t\bigg]_\pi^{3\pi/2}$
On applying limits
$= \large\frac{1}{\pi^2} $$\bigg[2( \sin \pi-\pi \cos \pi)-0\bigg]-\bigg[\bigg( \sin \large\frac{3\pi}{2}-\frac{3\pi}{2} $$\cos \large\frac{3\pi}{2}\bigg)$$-\bigg(\sin \pi-\pi \cos \pi\bigg)\bigg]$
$\sin \pi=0, \cos \pi=-1, \sin 3\pi/2=-1, \cos 3 \pi/2 =0$
$=\large\frac{1}{\pi^2} $$[2 \pi -(1-\pi)]$
$=\large\frac{1}{\pi^2}$$(3\pi+1)$
$=\large\frac{3}{\pi}+\frac{1}{\pi^2}$

 

answered Apr 27, 2013 by meena.p
edited Jan 9, 2014 by balaji.thirumalai
 
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