integral from 0 to x of (4-t^2)e^{t^2}

解

\int_{0}^{x} (4 - t^{2}) e^{t^{2}} d t

2 π erfi (x) - \frac{x e ^{x^{2}}}{2} + \frac{π}{4} erfi (x)

解答ステップ

\int_{0}^{x} (4 - t^{2}) e^{t^{2}} d t

拡張

(4 - t^{2}) e^{t^{2}} : 4 e^{t^{2}} - e^{t^{2}} t^{2}

= \int_{0}^{x} 4 e^{t^{2}} - e^{t^{2}} t^{2} d t

総和規則を適用する:

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

= \int_{0}^{x} 4 e^{t^{2}} d t - \int_{0}^{x} e^{t^{2}} t^{2} d t

\int_{0}^{x} 4 e^{t^{2}} d t = 2 π erfi (x)

\int_{0}^{x} e^{t^{2}} t^{2} d t = \frac{x e ^{x^{2}}}{2} - \frac{π}{4} erfi (x)

= 2 π erfi (x) - (\frac{x e ^{x^{2}}}{2} - \frac{π}{4} erfi (x))

簡素化

= 2 π erfi (x) - \frac{x e ^{x^{2}}}{2} + \frac{π}{4} erfi (x)