integral from 1 to 2 of xsin(4x)

解

\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}))

十進法表記

0.00004 \dots

解答ステップ

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

不定積分を計算する:

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

限度を計算する:

\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}))