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23 votes
3.3k views

The value of the integral given below is
$$\int \limits_0^{\pi} \: x^2 \: \cos x\:dx$$

  1. $-2\pi$
  2. $\pi$
  3. $-\pi$
  4. $2\pi$
in Calculus 3.3k views

4 Answers

21 votes
 
Best answer

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
2
$\int u.v=u.v_{1}-u^1.v_{2}+u^2.v_{3}+.....$

where u and v are your original functions

$u^1=first\,derivative\,of\,u$

$v_1=first\,integral\,of\,v$

$u^1=2x,u^2=2$ $\, v_1=sinx,v_2=-cosx,v_3=-sinx$

it is clear that after term $u^2$ everything in integral would be 0.

so plugging in values we get

$\int x^2 cosx dx=[x^2sinx+2xcosx-2sinx]$

now find $[x^2sinx+2xcosx-2sinx]_0^{\pi}=-2\pi$
0

Meaning of below statement (this statement is written at end of above answer) is formula of "integration by parts" in words. Here "a" and "b" are two functions.

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

 

0

sir way we havn’t do direct intigartion like noraml uv ? 

4 votes

use integration by parts, you will get 2π

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
1
@srestha ,  $2\pi$ is missing with $cos \ x$.
0
ok, thanks :)
0
if it would have asked as numerical type what value of pi should we take ??????

3.14 or radian 180
1

@priyanka manwani In 2018 they had specifically mentioned in the question to take it as 3.14.

4 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

I was going to post this method as an answer. Nice :-)

For those wondering, the method is from Oliver Knill's notes: http://people.math.harvard.edu/~knill/teaching/math1a_2012/handouts/39-parts.pdf.

0

@pritishc

Thanks for providing the pdf.

1
Thanks for this amazing method. Will save hell lot of time.
Answer:

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