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人気のある微分積分 >

integral from 2 to 9 of 2x(x-1)^{1/3}

解

\int_{2}^{9} 2 x (x - 1)^{\frac{1}{3}} d x

解

\frac{1839}{14}

+1

十進法表記

131.35714 \dots

解答ステップ

\int_{2}^{9} 2 x (x - 1)^{\frac{1}{3}} d x

定数を除く:

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

= 2 \cdot \int_{2}^{9} x (x - 1)^{\frac{1}{3}} d x

u置換積分法を適用する

= 2 \cdot \int_{1}^{2} 3 u^{3} (u^{3} + 1) d u

定数を除く:

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

= 2 \cdot 3 \cdot \int_{1}^{2} u^{3} (u^{3} + 1) d u

部分積分を適用する

= 2 \cdot 3 [\frac{1}{4} u^{4} (u^{3} + 1) - \int \frac{3 u ^{6}}{4} d u]_{1}^{2}

\int \frac{3 u ^{6}}{4} d u = \frac{3 u ^{7}}{28}

= 2 \cdot 3 [\frac{1}{4} u^{4} (u^{3} + 1) - \frac{3 u ^{7}}{28}]_{1}^{2}

簡素化

2 \cdot 3 [\frac{1}{4} u^{4} (u^{3} + 1) - \frac{3 u ^{7}}{28}]_{1}^{2} : 6 [\frac{1}{28} (4 u^{7} + 7 u^{4})]_{1}^{2}

= 6 [\frac{1}{28} (4 u^{7} + 7 u^{4})]_{1}^{2}

限度を計算する:

\frac{613}{28}

= 6 \cdot \frac{613}{28}

簡素化

= \frac{1839}{14}

グラフ

人気の例

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\int_{- 1}^{2} x^{2} + 4 d x

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\int_{0}^{4} x^{2} 2 x + 1 d x

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\int_{a}^{b} e^{- x^{2}} d x