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

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

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

\int_{0}^{π} x^{2} sin (2 x) d x

- \frac{π ^{2}}{2}

Decimal Notation

- 4.93480 \dots

Solution steps

\int_{0}^{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}^{π}

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

Simplify

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

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

Compute the boundaries:

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

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

Simplify

= - \frac{π ^{2}}{2}

\frac{d}{d x} (a x^{2} + b - 4)

\frac{2 s + 1}{s ^{3} + 7 s ^{2} + 12 s}

\frac{d}{d x} (cosh (0))

e^{x}, x^{2} - 1, x = - 1, x = 1

\int 29 sec^{- 2} (x t a) n^{3} x

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