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4 Answers

Best answer
26 votes
26 votes

Answer is $A$.

$\int_{0}^{\pi } x^{^{2}} \cos x dx$

$= x^2 \sin x ]_0^{\pi} - \int_0^{\pi} 2x \sin x$

$=  x^{^{2}} \sin x ]_0^{\pi} + 2x \cos x ]_0^{\pi}- \int_0^{\pi} 2 \cos x dx$

$= x^{^{2}} \sin x ]_0^{\pi} + 2x \cos x ]_0^{\pi}-  2 \sin x ]_0^{\pi}$

$=[\pi ^2 (0) -0] + 2[ \pi (-1)-0] -2[0-0]$

$=-2\pi$

Integral of a multiplied by b equals a multiplied by integral of b

minus

integral of derivative of a multiplied by integral of b

edited by
14 votes
14 votes

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

$I = -2\pi\:\:\:\:\:\:[\because\: \sin\pi = 0,\cos\pi = -1,\sin 0 = 0,\cos 0 = 1]$

So, the correct answer is $(A).$

4 votes
4 votes
$I=\int_{0}^{\pi }x^{2}cosx dx$............................................i

$I=\int_{0}^{\pi }\left ( \pi -x \right )^{2}cos\left ( \pi -x \right ) dx$

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

   $=-\int_{0}^{\pi }\left ( \pi^{2}-2\pi x+x^{2} \right )cosx dx$.....................................ii

Now adding i and ii we get

$2I=-\int_{0}^{\pi }\left ( \pi^{2}-2\pi x \right )cosx dx$

    $=-\pi ^{2}\int_{0}^{\pi}cosx dx$$+2\pi \int_{0}^{\pi}xcosx dx$

     $=- \pi ^{2}sinx$$+2\pi xsinx-\int_{0}^{\pi }sinxdx$

      $=2\pi\left [ -cosx \right ]_{0}^{\pi }$

      $=-4\pi$

$I=-2\pi$
edited by
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