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

解答

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

\frac{e ^{5} π}{2} erfi (2 x)

求解步骤

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

化简

e^{t^{2} + 5} : e^{t^{2}} e^{5}

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

提出常数:

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

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

这是非初等函数积分:

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

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

带入上下限计算后得:

\frac{π}{2} erfi (2 x)

= e^{5} \frac{π}{2} erfi (2 x)

化简

= \frac{e ^{5} π}{2} erfi (2 x)

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