integral from 0 to t of 2e^{2x}

解答

\int_{0}^{t} 2 e^{2 x} d x

e^{2 t} - 1

求解步骤

\int_{0}^{t} 2 e^{2 x} d x

提出常数:

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

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

使用换元积分法

= 2 \cdot \int_{0}^{2 t} 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

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

使用积分常见定则:

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

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

化简

2 \cdot \frac{1}{2} [e^{u}]_{0}^{2 t} : [e^{u}]_{0}^{2 t}

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

带入上下限计算后得:

e^{2 t} - 1

= e^{2 t} - 1

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