23 votes

The value of the integral given below is

$$\int \limits_0^{\pi} \: x^2 \: \cos x\:dx$$

- $-2\pi$
- $\pi$
- $-\pi$
- $2\pi$

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

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

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

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$

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

$=-4\pi$

$I=-2\pi$

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

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

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

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

$=-4\pi$

$I=-2\pi$

0

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.