integral from 0 to ln(A) of 6e^{2x}

解

\int_{0}^{l n (A)} 6 e^{2 x} d x

3 (A^{2} - 1)

解答ステップ

\int_{0}^{l n (A)} 6 e^{2 x} d x

定数を除く:

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

= 6 \cdot \int_{0}^{l n (A)} e^{2 x} d x

u置換積分法を適用する

= 6 \cdot \int_{0}^{2 l n (A)} 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

= 6 \cdot \frac{1}{2} \cdot \int_{0}^{2 l n (A)} e^{u} d u

共通積分を使用する:

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

= 6 \cdot \frac{1}{2} [e^{u}]_{0}^{2 l n (A)}

簡素化

6 \cdot \frac{1}{2} [e^{u}]_{0}^{2 l n (A)} : 3 [e^{u}]_{0}^{2 l n (A)}

= 3 [e^{u}]_{0}^{2 l n (A)}

限度を計算する:

A^{2} - 1

= 3 (A^{2} - 1)