ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral from 0 to 1 of x^8(1-x^3)^{5/4}

解

\int_{0}^{1} x^{8} (1 - x^{3})^{\frac{5}{4}} d x

解

\frac{128}{5967}

+1

十進法表記

0.02145 \dots

解答ステップ

\int_{0}^{1} x^{8} (1 - x^{3})^{\frac{5}{4}} d x

u置換積分法を適用する

= \int_{1}^{0} - \frac{u ^{\frac{5}{4}} ( - u + 1 ) ^{2}}{3} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - \frac{u ^{\frac{5}{4}} ( - u + 1 ) ^{2}}{3} d u

定数を除く:

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

= - (- \frac{1}{3} \cdot \int_{0}^{1} u^{\frac{5}{4}} (- u + 1)^{2} d u)

拡張

u^{\frac{5}{4}} (- u + 1)^{2} : u^{\frac{13}{4}} - 2 u^{\frac{9}{4}} + u^{\frac{5}{4}}

= - (- \frac{1}{3} \cdot \int_{0}^{1} u^{\frac{13}{4}} - 2 u^{\frac{9}{4}} + u^{\frac{5}{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

= - (- \frac{1}{3} (\int_{0}^{1} u^{\frac{13}{4}} d u - \int_{0}^{1} 2 u^{\frac{9}{4}} d u + \int_{0}^{1} u^{\frac{5}{4}} d u))

\int_{0}^{1} u^{\frac{13}{4}} d u = \frac{4}{17}

\int_{0}^{1} 2 u^{\frac{9}{4}} d u = \frac{8}{13}

\int_{0}^{1} u^{\frac{5}{4}} d u = \frac{4}{9}

= - (- \frac{1}{3} (\frac{4}{17} - \frac{8}{13} + \frac{4}{9}))

簡素化

- (- \frac{1}{3} (\frac{4}{17} - \frac{8}{13} + \frac{4}{9})) : \frac{128}{5967}

= \frac{128}{5967}

グラフ

人気の例

integral from-2 to 2 of xsqrt(2-x)

\int_{- 2}^{2} x 2 - x d x

integral from 0 to 1 of 10e^{-st}

\int_{0}^{1} 10 e^{- s t} d t

integral from 0 to 1 of e^{7x}

\int_{0}^{1} e^{7 x} d x

integral from-4 to 2 of (-x+2)

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

integral from-3 to 1 of 6x^2-5x+2

\int_{- 3}^{1} 6 x^{2} - 5 x + 2 d x