What is the integral from-pi to pi of (x^2+x)sin(x) ?

The integral from-pi to pi of (x^2+x)sin(x) is 2pi

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integral from-pi to pi of (x^2+x)sin(x)

\int_{- π}^{π} (x^{2} + x) sin (x) d x

2 π

Decimal Notation

6.28318 \dots

Solution steps

\int_{- π}^{π} (x^{2} + x) sin (x) d x

Apply Integration By Parts

: [- cos (x) (x^{2} + x) - \int - cos (x) (2 x + 1) d x]_{- π}^{π}

= [- cos (x) (x^{2} + x) - \int - cos (x) (2 x + 1) d x]_{- π}^{π}

\int - cos (x) (2 x + 1) d x = - 2 x sin (x) - sin (x) - 2 cos (x)

= [- cos (x) (x^{2} + x) - (- 2 x sin (x) - sin (x) - 2 cos (x))]_{- π}^{π}

Expand

- cos (x) (x^{2} + x) - (- 2 x sin (x) - sin (x) - 2 cos (x)) : - x^{2} cos (x) - x cos (x) + 2 x sin (x) + sin (x) + 2 cos (x)

= [- x^{2} cos (x) - x cos (x) + 2 x sin (x) + sin (x) + 2 cos (x)]_{- π}^{π}

Compute the boundaries:

2 π

= 2 π

\int_{1}^{2} \frac{x}{3 x ^{2} - 2} d x

\int_{- 1}^{0} 5 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

\int_{1}^{4} \frac{1}{x + 1 + x} d x

\int_{- 5}^{- 1} \frac{1}{16} d x

\int_{0}^{\frac{π}{2}} e^{- s t} sin (t) d t

What is the integral from-pi to pi of (x^2+x)sin(x) ?
The integral from-pi to pi of (x^2+x)sin(x) is 2pi