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

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

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

\int_{0}^{2 π} (x^{2} sin (2 x)) d x

- 2 π^{2}

Decimal Notation

- 19.73920 \dots

Solution steps

\int_{0}^{2 π} x^{2} sin (2 x) d x

Apply Integration By Parts

: [\frac{1}{2} (- x^{2} cos (2 x) - 2 \cdot \int - x cos (2 x) d x)]_{0}^{2 π}

= [\frac{1}{2} (- x^{2} cos (2 x) - 2 \cdot \int - x cos (2 x) d x)]_{0}^{2 π}

\int - x cos (2 x) d x = - \frac{1}{4} (2 x sin (2 x) + cos (2 x))

= [\frac{1}{2} (- x^{2} cos (2 x) - 2 (- \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{2 π}

Simplify

[\frac{1}{2} (- x^{2} cos (2 x) - 2 (- \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{2 π} : [\frac{1}{4} (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))]_{0}^{2 π}

= [\frac{1}{4} (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))]_{0}^{2 π}

Compute the boundaries:

\frac{- 8 π ^{2} + 1}{4} - \frac{1}{4}

= \frac{- 8 π ^{2} + 1}{4} - \frac{1}{4}

Simplify

= - 2 π^{2}

t \to 0 lim (\frac{sin ( 7 t )}{t})

\frac{d}{d x} (cos (x) + i sin (x))

y = x^{3} - 3 x + 2, (2, 4)

\int_{- \infty}^{- 1} e^{- 2 x} d x

\int \frac{3 x ^{3} - 3 x ^{2} + 1}{x ^{2} - x} d x

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