integral from 0 to 4 of e^{0.5x}

解答

\int_{0}^{4} e^{0.5 x} d x

2 (e^{2} - 1)

十进制

12.77811 \dots

求解步骤

\int_{0}^{4} e^{0.5 x} d x

使用换元积分法

= \int_{0}^{2} e^{u} \frac{1}{0.5} d u

提出常数:

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

= \frac{1}{0.5} \cdot \int_{0}^{2} e^{u} d u

化简

= 2 \cdot \int_{0}^{2} e^{u} d u

使用积分常见定则:

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

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

带入上下限计算后得:

e^{2} - 1

= 2 (e^{2} - 1)

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