What is the integral from 1 to 2 of xsin(4x) ?

The integral from 1 to 2 of xsin(4x) is

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integral from 1 to 2 of xsin(4x)

\int_{1^{\circ}}^{2^{\circ}} x sin (4 x) d x

- \frac{1}{4} \cdot 2^{\circ} cos (4 \cdot 2^{\circ}) + \frac{1}{16} sin (4 \cdot 2^{\circ}) - (- \frac{1}{4} \cdot 1^{\circ} cos (4 \cdot 1^{\circ}) + \frac{1}{16} sin (4 \cdot 1^{\circ}))

Decimal Notation

0.00004 \dots

Solution steps

\int_{1^{\circ}}^{2^{\circ}} x sin (4 x) d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{1^{\circ}}^{2^{\circ}} x sin (4 x) d x = - \frac{1}{4} \cdot 2^{\circ} cos (4 \cdot 2^{\circ}) + \frac{1}{16} sin (4 \cdot 2^{\circ}) - (- \frac{1}{4} \cdot 1^{\circ} cos (4 \cdot 1^{\circ}) + \frac{1}{16} sin (4 \cdot 1^{\circ}))

= - \frac{1}{4} \cdot 2^{\circ} cos (4 \cdot 2^{\circ}) + \frac{1}{16} sin (4 \cdot 2^{\circ}) - (- \frac{1}{4} \cdot 1^{\circ} cos (4 \cdot 1^{\circ}) + \frac{1}{16} sin (4 \cdot 1^{\circ}))

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What is the integral from 1 to 2 of xsin(4x) ?
The integral from 1 to 2 of xsin(4x) is