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人気のある微分積分 >

integral from 0 to pi of x^3cos(2x)

解

\int_{0}^{π} x^{3} cos (2 x) d x

解

\frac{3 π ^{2}}{4}

+1

十進法表記

7.40220 \dots

解答ステップ

\int_{0}^{π} x^{3} cos (2 x) d x

部分積分を適用する

= [\frac{1}{2} (x^{3} sin (2 x) - 2 \cdot \int \frac{3}{2} x^{2} sin (2 x) d x)]_{0}^{π}

\int \frac{3}{2} x^{2} sin (2 x) d x = \frac{3}{2} (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x)))

= [\frac{1}{2} (x^{3} sin (2 x) - 2 \cdot \frac{3}{2} (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{π}

簡素化

[\frac{1}{2} (x^{3} sin (2 x) - 2 \cdot \frac{3}{2} (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{π} : [\frac{1}{2} (x^{3} sin (2 x) - 3 (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{π}

= [\frac{1}{2} (x^{3} sin (2 x) - 3 (- \frac{1}{2} x^{2} cos (2 x) + \frac{1}{4} (2 x sin (2 x) + cos (2 x))))]_{0}^{π}

限度を計算する:

- \frac{3 ( 1 - 2 π ^{2} )}{8} + \frac{3}{8}

= - \frac{3 ( 1 - 2 π ^{2} )}{8} + \frac{3}{8}

簡素化

= \frac{3 π ^{2}}{4}

グラフ

人気の例

integral from 2 to infinity of x^{-1/3}

\int_{2}^{\infty} x^{- \frac{1}{3}} d x

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\int_{- 1}^{2} (- x^{4} - x^{2} + 6 x + 8) d x

integral from 0 to 2pi of sin(2x)sin(3x)

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\int_{0}^{2} 7 x^{4} d x

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

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