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人気のある微分積分 >

integral from 1 to 2 of 1/(3-z)

解

\int_{1}^{2} \frac{1}{3 - z} d z

解

ln (2)

+1

十進法表記

0.69314 \dots

解答ステップ

\int_{1}^{2} \frac{1}{3 - z} d z

u置換積分法を適用する

= \int_{2}^{1} - \frac{1}{u} d u

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

= - \int_{1}^{2} - \frac{1}{u} d u

定数を除く:

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

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

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{1}^{2})

簡素化

= [ln ∣ u ∣]_{1}^{2}

限度を計算する:

ln (2)

= ln (2)

グラフ

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