integral from 0 to ln(A) of 4e^{2x}

解答

\int_{0}^{l n (A)} 4 e^{2 x} d x

2 (A^{2} - 1)

求解步骤

\int_{0}^{l n (A)} 4 e^{2 x} d x

提出常数:

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

= 4 \cdot \int_{0}^{l n (A)} e^{2 x} d x

使用换元积分法

= 4 \cdot \int_{0}^{2 l n (A)} e^{u} \frac{1}{2} d u

提出常数:

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

= 4 \cdot \frac{1}{2} \cdot \int_{0}^{2 l n (A)} e^{u} d u

使用积分常见定则:

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

= 4 \cdot \frac{1}{2} [e^{u}]_{0}^{2 l n (A)}

化简

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

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

带入上下限计算后得:

A^{2} - 1

= 2 (A^{2} - 1)

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