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 12 of 9/(sqrt(12-x))

解

\int_{0}^{12} \frac{9}{12 - x} d x

解

363

+1

十進法表記

62.35382 \dots

解答ステップ

\int_{0}^{12} \frac{9}{12 - x} 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}^{12} \frac{1}{12 - x} d x

u置換積分法を適用する

= 9 \cdot \int_{12}^{0} - \frac{1}{u} d u

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

= 9 (- \int_{0}^{12} - \frac{1}{u} d u)

定数を除く:

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

= 9 (- (- \int_{0}^{12} \frac{1}{u} d u))

べき乗則を適用する

= 9 (- (- [2 u^{\frac{1}{2}}]_{0}^{12}))

簡素化

9 (- (- [2 u^{\frac{1}{2}}]_{0}^{12})) : 9 [2 u]_{0}^{12}

= 9 [2 u]_{0}^{12}

限度を計算する:

43

= 9 \cdot 43

簡素化

= 363

グラフ

人気の例

integral from 0 to 1 of (e^x+x^e)

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

integral from 0 to 1 of (8t)/(t^2+1)

\int_{0}^{1} \frac{8 t}{t ^{2} + 1} d t

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

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

integral from-1 to 0 of x^2sqrt(1+x)

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

integral from 0 to 2 of e^{t^2}

\int_{0}^{2} e^{t^{2}} d t