integral from 0 to s of 1/(0.2s+10)

解

\int_{0}^{s} \frac{1}{0.2 s + 10} d s

5 (ln ∣ 0.2 s + 10 ∣ - 2.30258 \dots)

解答ステップ

\int_{0}^{s} \frac{1}{0.2 s + 10} d s

u置換積分法を適用する

= \int_{10}^{0.2 s + 10} \frac{1}{0.2 u} d u

定数を除く:

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

= \frac{1}{0.2} \cdot \int_{10}^{0.2 s + 10} \frac{1}{u} d u

簡素化

= 5 \cdot \int_{10}^{0.2 s + 10} \frac{1}{u} d u

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 [ln ∣ u ∣]_{10}^{0.2 s + 10}

限度を計算する:

ln ∣ 0.2 s + 10 ∣ - 2.30258 \dots

= 5 (ln ∣ 0.2 s + 10 ∣ - 2.30258 \dots)