# GATE2014-3-47

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

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
0
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 ?

use integration by parts, you will get 2π

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

1

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

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

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

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.

## Related questions

1
4.9k views
If $\int \limits_0^{2 \pi} |x \: \sin x| dx=k\pi$, then the value of $k$ is equal to ______.
1 vote
2
386 views
Example of one Question for preparing exam: Fourier series of function: be like as: $f(x)=\frac{a_0}{2}+\Sigma_{n=1}^{\infty} (a_n \cos nx+b_n \sin nx)$ (Question ) so the coefficient is: $a_n=0,n=2k+1,b_n=0,n=2k$ I want to find that how the coefficient is solved, this is my ... I think my solution is wrong, anyone could help me? I so sad...
3
3.8k views
If for non-zero $x, \: af(x) + bf(\frac{1}{x}) = \frac{1}{x} - 25$ where a $a \neq b \text{ then } \int_1^2 f(x)dx$ is $\frac{1}{a^2 - b^2} \begin{bmatrix} a(\ln 2 - 25) + \frac{47b}{2} \end{bmatrix}$ ... $\frac{1}{a^2 - b^2} \begin{bmatrix} a(\ln 2 - 25) - \frac{47b}{2} \end{bmatrix}$
Let $S = \sum_{i=3}^{100} i \log_{2} i$, and $T = \int_{2}^{100} x \log_{2}x dx$. Which of the following statements is true? $S > T$ $S = T$ $S < T$ and $2S > T$ $2S ≤ T$