integral from 0 to t of (t-x)^3e^x

解

\int_{0}^{t} (t - x)^{3} e^{x} d x

- t^{3} - 3 t^{2} - 6 t + 6 e^{t} - 6

解答ステップ

\int_{0}^{t} (t - x)^{3} e^{x} d x

拡張

(t - x)^{3} e^{x} : t^{3} e^{x} - 3 t^{2} e^{x} x + 3 t e^{x} x^{2} - e^{x} x^{3}

= \int_{0}^{t} t^{3} e^{x} - 3 t^{2} e^{x} x + 3 t e^{x} x^{2} - e^{x} x^{3} d x

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}^{t} t^{3} e^{x} d x - \int_{0}^{t} 3 t^{2} e^{x} x d x + \int_{0}^{t} 3 t e^{x} x^{2} d x - \int_{0}^{t} e^{x} x^{3} d x

\int_{0}^{t} t^{3} e^{x} d x = t^{3} (e^{t} - 1)

\int_{0}^{t} 3 t^{2} e^{x} x d x = 3 t^{2} (t e^{t} - e^{t} + 1)

\int_{0}^{t} 3 t e^{x} x^{2} d x = 3 t (t^{2} e^{t} - 2 (e^{t} t - e^{t}) - 2)

\int_{0}^{t} e^{x} x^{3} d x = t^{3} e^{t} - 3 (t^{2} e^{t} - 2 (t e^{t} - e^{t})) + 6

= t^{3} (e^{t} - 1) - 3 t^{2} (t e^{t} - e^{t} + 1) + 3 t (t^{2} e^{t} - 2 (e^{t} t - e^{t}) - 2) - (t^{3} e^{t} - 3 (t^{2} e^{t} - 2 (t e^{t} - e^{t})) + 6)

簡素化

t^{3} (e^{t} - 1) - 3 t^{2} (t e^{t} - e^{t} + 1) + 3 t (t^{2} e^{t} - 2 (e^{t} t - e^{t}) - 2) - (t^{3} e^{t} - 3 (t^{2} e^{t} - 2 (t e^{t} - e^{t})) + 6) : - t^{3} - 3 t^{2} - 6 t + 6 e^{t} - 6

= - t^{3} - 3 t^{2} - 6 t + 6 e^{t} - 6