integral from 0 to x of e^{5t^2}

解答

\int_{0}^{x} e^{5 t^{2}} d t

\frac{π}{2 5} erfi (5 x)

求解步骤

\int_{0}^{x} e^{5 t^{2}} d t

= \int_{0}^{x} e^{(5 t)^{2}} d t

使用换元积分法

= \int_{0}^{5 x} e^{u^{2}} \frac{1}{5} d u

提出常数:

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

= \frac{1}{5} \cdot \int_{0}^{5 x} e^{u^{2}} d u

这是非初等函数积分:

\int e^{u^{2}} d u = \frac{π}{2} erfi (u)

= \frac{1}{5} [\frac{π}{2} erfi (u)]_{0}^{5 x}

带入上下限计算后得:

\frac{π}{2} erfi (5 x)

= \frac{1}{5} \cdot \frac{π}{2} erfi (5 x)

化简

= \frac{π}{2 5} erfi (5 x)

\int_{0}^{1} (5 x + x) d x

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