integral from 0 to 2 of (xe^{x^2})

解答

\int_{0}^{2} (x e^{x^{2}}) d x

\frac{e ^{4} - 1}{2}

十进制

26.79907 \dots

求解步骤

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

使用换元积分法

= \int_{0}^{4} \frac{e ^{u}}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int_{0}^{4} e^{u} d u

使用积分常见定则:

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

= \frac{1}{2} [e^{u}]_{0}^{4}

带入上下限计算后得:

e^{4} - 1

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

化简

= \frac{e ^{4} - 1}{2}

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