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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 | 684 views

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

edited by

use integration by parts, you will get 2π

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