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人気のある微分積分 >

integral from 0 to 1 of e^{-5t}

解

\int_{0}^{1} e^{- 5 t} d t

解

\frac{e ^{5} - 1}{5 e ^{5}}

+1

十進法表記

0.19865 \dots

解答ステップ

\int_{0}^{1} e^{- 5 t} d t

u置換積分法を適用する

= \int_{0}^{- 5} - \frac{1}{5} e^{u} d u

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

= - \int_{- 5}^{0} - \frac{1}{5} e^{u} d u

定数を除く:

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

= - (- \frac{1}{5} \cdot \int_{- 5}^{0} e^{u} d u)

共通積分を使用する:

\int e^{u} d u = e^{u}

= - (- \frac{1}{5} [e^{u}]_{- 5}^{0})

簡素化

= \frac{1}{5} [e^{u}]_{- 5}^{0}

限度を計算する:

1 - \frac{1}{e ^{5}}

= \frac{1}{5} (1 - \frac{1}{e ^{5}})

簡素化

= \frac{e ^{5} - 1}{5 e ^{5}}

グラフ

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