integral from t to 0 of e^{ax}

解

\int_{t}^{0} e^{a x} d x

\frac{1 - e ^{a t}}{a}

解答ステップ

\int_{t}^{0} e^{a x} d x

u置換積分法を適用する

= \int_{a t}^{0} \frac{e ^{u}}{a} d u

定数を除く:

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

= \frac{1}{a} \cdot \int_{a t}^{0} e^{u} d u

共通積分を使用する:

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

= \frac{1}{a} [e^{u}]_{a t}^{0}

簡素化

\frac{1}{a} [e^{u}]_{a t}^{0} : \frac{[ e ^{u} ] _{a t}^{0}}{a}

= \frac{[ e ^{u} ] _{a t}^{0}}{a}

限度を計算する:

1 - e^{a t}

= \frac{1 - e ^{a t}}{a}