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 sqrt(1-\sqrt{x)}

解

\int_{0}^{1} 1 - x d x

解

\frac{8}{15}

+1

十進法表記

0.53333 \dots

解答ステップ

\int_{0}^{1} 1 - x d x

u置換積分法を適用する

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

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

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

定数を除く:

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

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

拡張

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

= - (- 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

= - (- 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}

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

簡素化

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

= \frac{8}{15}

グラフ

人気の例

integral from-1 to 1 of (1/(x^2))

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

integral from 0 to 1 of 2tsqrt(4t^2+1)

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

integral from 1 to 8 of (3x+2)

\int_{1}^{8} (3 x + 2) d x

integral from 1 to x of 1/(t^2+1)

\int_{1}^{x} \frac{1}{t ^{2} + 1} d t

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

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