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The value of the Integral $I = \displaystyle{}\int_{0}^{\pi/2} x^{2}\sin x dx$ is

  1. $(x+2)/2$
  2. $2/(\pi-2)$
  3. $\pi – 2$
  4. $\pi + 2$
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We will use the method of integration by parts. Using the $ILATE$ rule, we choose  $x^{2}$  as first function and  $sin(x)$  as the second function.

By the definition of integration by parts,  $\int I\times II$  $dx$ $=$  $I\times\int II$ $dx$  $-$ $\int (\frac{d}{dx}(I)\times \int IIdx)$  $dx$     (Reference:  Integration by parts)

Let the answer de denoted by  $I_{1}$

$\therefore$ $\int x^{2}sin(x)dx=x^{2}\times \int sin(x)dx-\int 2x(-cos(x))dx$   ----(1)

 

Now, $\int xcos(x)dx=xsin(x)-\int sinxdx$  $=$   $xsin(x)+cos(x)$  -----(2)

 

Putting the value of (2) in (1), we get

 

$\therefore$ $\int x^{2}sin(x)dx=x^{2}(-cos(x))+2(xsin(x)+cos(x))$  $=$  $cos(x)(2-x^{2})+2xsin(x)$

 

Substituting the limits, we get :  $I_{1}=\left [ (0+\pi)-(2+0) \right ]=\pi-2$

Option C is correct.

Answer:

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