What is the integral from 0 to pi of 2x^2sin(1/2 x) ?

The integral from 0 to pi of 2x^2sin(1/2 x) is 2(8pi-16)

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integral from 0 to pi of 2x^2sin(1/2 x)

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

2 (8 π - 16)

Decimal Notation

18.26548 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{π} x^{2} sin (\frac{1}{2} x) d x

Simplify

\frac{1}{2} x : \frac{x}{2}

= 2 \cdot \int_{0}^{π} x^{2} sin (\frac{x}{2}) d x

Apply Integration By Parts

: [- 2 x^{2} cos (\frac{x}{2}) - \int - 4 x cos (\frac{x}{2}) d x]_{0}^{π}

= 2 [- 2 x^{2} cos (\frac{x}{2}) - \int - 4 x cos (\frac{x}{2}) d x]_{0}^{π}

\int - 4 x cos (\frac{x}{2}) d x = - 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))

= 2 [- 2 x^{2} cos (\frac{x}{2}) - (- 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2})))]_{0}^{π}

Simplify

= 2 [- 2 x^{2} cos (\frac{x}{2}) + 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))]_{0}^{π}

Compute the boundaries:

8 π - 16

= 2 (8 π - 16)

\frac{4 x ^{2} + 2 x + 2}{x}

\frac{d}{d x} (x^{2} - 7 x + 2)

\frac{d}{d x} (\frac{( x + 1 - 1 )}{x})

\frac{d}{d x} (3 e^{x} + 5 x^{2})

3 \cdot \frac{d y}{d x} + 5 y = 1

What is the integral from 0 to pi of 2x^2sin(1/2 x) ?
The integral from 0 to pi of 2x^2sin(1/2 x) is 2(8pi-16)