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

解

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

\frac{π}{2} erfi (x^{2}) - \frac{1}{2} (e^{x^{4}} - 1)

解答ステップ

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

拡張

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

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

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

\int_{0}^{x^{2}} e^{t^{2}} t d t = \frac{1}{2} (e^{x^{4}} - 1)

= \frac{π}{2} erfi (x^{2}) - \frac{1}{2} (e^{x^{4}} - 1)