What is the integral from 0 to 2pi of x^2sin(x) ?

The integral from 0 to 2pi of x^2sin(x) is -4pi^2

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integral from 0 to 2pi of x^2sin(x)

\int_{0}^{2 π} x^{2} sin (x) d x

- 4 π^{2}

Decimal Notation

- 39.47841 \dots

Solution steps

\int_{0}^{2 π} x^{2} sin (x) d x

Apply Integration By Parts

: [- x^{2} cos (x) - \int - 2 x cos (x) d x]_{0}^{2 π}

= [- x^{2} cos (x) - \int - 2 x cos (x) d x]_{0}^{2 π}

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

= [- x^{2} cos (x) - (- 2 (x sin (x) + cos (x)))]_{0}^{2 π}

Simplify

= [- x^{2} cos (x) + 2 (x sin (x) + cos (x))]_{0}^{2 π}

Compute the boundaries:

- 4 π^{2}

= - 4 π^{2}

\int_{0}^{\frac{5 π}{6}} sin (x) d x

\int_{0}^{1} 8 cos (\frac{π x}{2}) d x

\int_{0}^{1} (x^{2} + x - 1) d x

\int_{0}^{2 π} cos (n x) d x

\int_{0}^{3 x} f (t) d t

What is the integral from 0 to 2pi of x^2sin(x) ?
The integral from 0 to 2pi of x^2sin(x) is -4pi^2