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 8x^3sqrt(1-x^2)

解

\int_{0}^{1} 8 x^{3} 1 - x^{2} d x

解

\frac{16}{15}

+1

十進法表記

1.06666 \dots

解答ステップ

\int_{0}^{1} 8 x^{3} 1 - x^{2} d x

定数を除く:

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

= 8 \cdot \int_{0}^{1} x^{3} 1 - x^{2} d x

u置換積分法を適用する

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

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

= 8 (- \int_{0}^{1} - \frac{( - u + 1 ) u}{2} d u)

定数を除く:

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

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

拡張

(- u + 1) u : - u^{\frac{3}{2}} + u

= 8 (- (- \frac{1}{2} \cdot \int_{0}^{1} - u^{\frac{3}{2}} + u 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

= 8 (- (- \frac{1}{2} (- \int_{0}^{1} u^{\frac{3}{2}} d u + \int_{0}^{1} u d u)))

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

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

= 8 (- (- \frac{1}{2} (- \frac{2}{5} + \frac{2}{3})))

簡素化

8 (- (- \frac{1}{2} (- \frac{2}{5} + \frac{2}{3}))) : \frac{16}{15}

= \frac{16}{15}

グラフ

人気の例

integral from \sqrt[3]{x} to pi of sin(t^3)

\int_{3 x}^{π} sin (t^{3}) d t

integral from 0 to pi/3 of 2cos(x)+1

\int_{0}^{\frac{π}{3}} 2 cos (x) + 1 d x

integral from 0 to 1 of (10^{3x}+10^{-3x})

\int_{0}^{1} (1 0^{3 x} + 1 0^{- 3 x}) d x

integral from 0 to 4 of g(x)

\int_{0}^{4} g (x) d x

integral from 0 to 9 of 9-x

\int_{0}^{9} 9 - x d x