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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. $2\pi$
asked in Calculus by Veteran (87.2k points)   | 684 views

2 Answers

+9 votes
Best answer

ans 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

answered by Boss (6.3k points)  
edited by
+3 votes

use integration by parts, you will get 2π

answered by Veteran (13.4k points)  


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