integral of (9-t^2)e^{t^2}

Lösung

\int (9 - t^{2}) e^{t^{2}} d t

9 \cdot \frac{π}{2} erfi (t) - \frac{1}{2} e^{t^{2}} t + \frac{π}{4} erfi (t) + C

Schritte zur Lösung

\int (9 - t^{2}) e^{t^{2}} d t

Multipliziere aus

(9 - t^{2}) e^{t^{2}} : 9 e^{t^{2}} - e^{t^{2}} t^{2}

= \int 9 e^{t^{2}} - e^{t^{2}} t^{2} d t

Wende die Summenregel an:

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

= \int 9 e^{t^{2}} d t - \int e^{t^{2}} t^{2} d t

\int 9 e^{t^{2}} d t = 9 \cdot \frac{π}{2} erfi (t)

\int e^{t^{2}} t^{2} d t = \frac{1}{2} e^{t^{2}} t - \frac{π}{4} erfi (t)

= 9 \cdot \frac{π}{2} erfi (t) - (\frac{1}{2} e^{t^{2}} t - \frac{π}{4} erfi (t))

Vereinfache

= 9 \cdot \frac{π}{2} erfi (t) - \frac{1}{2} e^{t^{2}} t + \frac{π}{4} erfi (t)

Füge eine Konstante zur Lösung hinzu

= 9 \cdot \frac{π}{2} erfi (t) - \frac{1}{2} e^{t^{2}} t + \frac{π}{4} erfi (t) + C

y^{^{'}} = 2 y + e^{2 x} cos (3 x)

\int y \cdot 25 \cdot (y - Θ) \cdot e^{(- 5 (y - Θ))} d y

\int y^{2} e^{y} d y

s l o p e (- 6, 6), (4, - 7)

3 y^{^{''}} = 2 y^{^{'}}