integral from 1 to 3 of xe^{2x}

解答

\int_{1}^{3} x e^{2 x} d x

\frac{5 e ^{6} - e ^{2}}{4}

十进制

502.43872 \dots

求解步骤

\int_{1}^{3} x e^{2 x} d x

使用换元积分法

= \int_{2}^{6} \frac{e ^{u} u}{4} d u

提出常数:

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

= \frac{1}{4} \cdot \int_{2}^{6} e^{u} u d u

使用分布积分法

= \frac{1}{4} [e^{u} u - \int e^{u} d u]_{2}^{6}

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

= \frac{1}{4} [e^{u} u - e^{u}]_{2}^{6}

带入上下限计算后得:

5 e^{6} - e^{2}

= \frac{1}{4} (5 e^{6} - e^{2})

化简

= \frac{5 e ^{6} - e ^{2}}{4}

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