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 log_{e}(x)

解

\int_{1}^{2} lo g_{e} (x) d x

解

2 ln (2) - 1

+1

十進法表記

0.38629 \dots

解答ステップ

\int_{1}^{2} lo g_{e} (x) d x

部分積分を適用する

= [x ln (x) - \int 1 d x]_{1}^{2}

\int 1 d x = x

= [x ln (x) - x]_{1}^{2}

限度を計算する:

2 ln (2) - 1

= 2 ln (2) - 1

グラフ

人気の例

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

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

integral from 0 to pi/2 of (e^t+cos(t))

\int_{0}^{\frac{π}{2}} (e^{t} + cos (t)) d t

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

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

integral from 0 to 1/2 of (2x+1)e^x

\int_{0}^{\frac{1}{2}} (2 x + 1) e^{x} d x

integral from 3 to 4 of 1/(25-x^2)

\int_{3}^{4} \frac{1}{25 - x ^{2}} d x