integral from-2 to 0 of (x+1)e^{x^2+2x}

Lösung

\int_{- 2}^{0} (x + 1) e^{x^{2} + 2 x} d x

- \frac{π erfi ( 1 ) - π erfi ( - 1 )}{2 e} + \frac{π erfi ( 1 ) - π erfi ( - 1 )}{2 e}

Schritte zur Lösung

\int_{- 2}^{0} (x + 1) e^{x^{2} + 2 x} d x

Multipliziere aus

(x + 1) e^{x^{2} + 2 x} : e^{x^{2} + 2 x} x + e^{x^{2} + 2 x}

= \int_{- 2}^{0} e^{x^{2} + 2 x} x + e^{x^{2} + 2 x} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 2}^{0} e^{x^{2} + 2 x} x d x + \int_{- 2}^{0} e^{x^{2} + 2 x} d x

\int_{- 2}^{0} e^{x^{2} + 2 x} x d x = - \frac{π erfi ( 1 ) - π erfi ( - 1 )}{2 e}

\int_{- 2}^{0} e^{x^{2} + 2 x} d x = \frac{π erfi ( 1 ) - π erfi ( - 1 )}{2 e}

= - \frac{π erfi ( 1 ) - π erfi ( - 1 )}{2 e} + \frac{π erfi ( 1 ) - π erfi ( - 1 )}{2 e}

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