integral from 0 to c of 10e^{2x}

解

\int_{0}^{c} 10 e^{2 x} d x

5 (e^{2 c} - 1)

解答ステップ

\int_{0}^{c} 10 e^{2 x} d x

定数を除く:

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

= 10 \cdot \int_{0}^{c} e^{2 x} d x

u置換積分法を適用する

= 10 \cdot \int_{0}^{2 c} e^{u} \frac{1}{2} d u

定数を除く:

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

= 10 \cdot \frac{1}{2} \cdot \int_{0}^{2 c} e^{u} d u

共通積分を使用する:

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

= 10 \cdot \frac{1}{2} [e^{u}]_{0}^{2 c}

簡素化

10 \cdot \frac{1}{2} [e^{u}]_{0}^{2 c} : 5 [e^{u}]_{0}^{2 c}

= 5 [e^{u}]_{0}^{2 c}

限度を計算する:

e^{2 c} - 1

= 5 (e^{2 c} - 1)