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人気のある微分積分 >

integral from 0 to 1 of 9x^4(8-x^5)^2

解

\int_{0}^{1} 9 x^{4} (8 - x^{5})^{2} d x

解

\frac{507}{5}

+1

十進法表記

101.4

解答ステップ

\int_{0}^{1} 9 x^{4} (8 - x^{5})^{2} d x

定数を除く:

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

= 9 \cdot \int_{0}^{1} x^{4} (8 - x^{5})^{2} d x

因数

8 - x^{5} : - (- 8 + x^{5})

= 9 \cdot \int_{0}^{1} x^{4} (- (- 8 + x^{5}))^{2} d x

(- (- 8 + x^{5}))^{2} = (- 8 + x^{5})^{2}

= 9 \cdot \int_{0}^{1} x^{4} (- 8 + x^{5})^{2} d x

u置換積分法を適用する

= 9 \cdot \int_{0}^{1} \frac{u ^{\frac{1}{4}} ( u ^{\frac{5}{4}} - 8 ) ^{2}}{4} d u

定数を除く:

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

= 9 \cdot \frac{1}{4} \cdot \int_{0}^{1} u^{\frac{1}{4}} (u^{\frac{5}{4}} - 8)^{2} d u

拡張

u^{\frac{1}{4}} (u^{\frac{5}{4}} - 8)^{2} : u^{\frac{11}{4}} - 16 u^{\frac{3}{2}} + 64 u^{\frac{1}{4}}

= 9 \cdot \frac{1}{4} \cdot \int_{0}^{1} u^{\frac{11}{4}} - 16 u^{\frac{3}{2}} + 64 u^{\frac{1}{4}} d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 \cdot \frac{1}{4} (\int_{0}^{1} u^{\frac{11}{4}} d u - \int_{0}^{1} 16 u^{\frac{3}{2}} d u + \int_{0}^{1} 64 u^{\frac{1}{4}} d u)

\int_{0}^{1} u^{\frac{11}{4}} d u = \frac{4}{15}

\int_{0}^{1} 16 u^{\frac{3}{2}} d u = \frac{32}{5}

\int_{0}^{1} 64 u^{\frac{1}{4}} d u = \frac{256}{5}

= 9 \cdot \frac{1}{4} (\frac{4}{15} - \frac{32}{5} + \frac{256}{5})

簡素化

9 \cdot \frac{1}{4} (\frac{4}{15} - \frac{32}{5} + \frac{256}{5}) : \frac{507}{5}

= \frac{507}{5}

グラフ

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