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 1 to 2 of ((2sqrt(x-1)))/x

解

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

解

- π + 4

+1

十進法表記

0.85840 \dots

解答ステップ

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

定数を除く:

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

= 2 \cdot \int_{1}^{2} \frac{x - 1}{x} d x

u置換積分法を適用する

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

定数を除く:

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

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

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

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

\int_{0}^{1} \frac{1}{u ^{2} + 1} d u = \frac{π}{4}

\int_{0}^{1} 1 d u = 1

= 2 \cdot 2 (- \frac{π}{4} + 1)

簡素化

2 \cdot 2 (- \frac{π}{4} + 1) : - π + 4

= - π + 4

グラフ

人気の例

inverselaplace 1/(s+4)

in v erse l a pl a ce \frac{1}{s + 4}

derivative of 4/((x+1^2))

\frac{d}{d x} (\frac{4}{( x + 1 ) ^{2}})

integral of (2/x-3/(x^4))

\int (\frac{2}{x} - \frac{3}{x ^{4}}) d x

s l o p e y = \frac{2}{x}

integral of x^{1/2}sqrt(1+x^{1/2)}

\int x^{\frac{1}{2}} 1 + x^{\frac{1}{2}} d x