GATE CSE 2014 | Set 3 | Question: 47

Question

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

• A. $-2\pi$
• B. $\pi$
• C. $-\pi$
• D. $2\pi$

Comments

A

Answer 1

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

Comments

for exam do in determinant and then change the sign we will get the answer.

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I didn't understand what you meant by determinant bro , can you explain please @pavansan

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@Nebula actually it wont work for variables but it will work for constants for constants if you do determinant and inverse then we will get integration as result try this method for constants.

Answer 2

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

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

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

Comments

This method is also well explained in this link along with proof.

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This only works where the first function is algebraic right? I think Sachin sir taught it in class.

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@Nebula yes it works if it is polynomial x exponential or polynomial x trignometric forms

Answer 3

use integration by parts, you will get −2π

Answer 4

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

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

      $=-4\pi$

$I=-2\pi$

Comments

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

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How did you replaced x with pi -x

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Actually there is a property  $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$   and we can prove it  easily –

lets suppose x= a+b-t  then dx= -dt  , t= a+b-x now put these values in above integration  

$-\int_{b}^{a} f(a+b-t) dt$ , and there is one property too.

$\int_{a}^{b} f(x) dx = - \int_{b}^{a} f(x) dx$ 

when we use this  will ultimately become 

$-\int_{b}^{a} f(a+b-t) dt = \int_{a}^{b} f(a+b-t) dt$

Answer 5

So the Correct Answer is (A).

Answer 6

Ans. : -2 pi