integral from 2x to x+5 of e^{t^2+1}

Lösung

\int_{2 x}^{x + 5} e^{t^{2} + 1} d t

e (\frac{π}{2} erfi (x + 5) - \frac{π}{2} erfi (2 x))

Schritte zur Lösung

\int_{2 x}^{x + 5} e^{t^{2} + 1} d t

e^{t^{2} + 1} = e^{1} e^{t^{2}}

= \int_{2 x}^{x + 5} e^{1} e^{t^{2}} d t

Entferne die Konstante:

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

= e^{1} \cdot \int_{2 x}^{x + 5} e^{t^{2}} d t

Vereinfache

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

Das ist ein nicht elementares Integral :

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

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

Berechne die Grenzen:

\frac{π}{2} erfi (x + 5) - \frac{π}{2} erfi (2 x)

= e (\frac{π}{2} erfi (x + 5) - \frac{π}{2} erfi (2 x))

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