integral from 0 to t of e^uu^2(t-u)^2

Lösung

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

2 t^{2} e^{t} - 2 t^{2} - 12 t e^{t} - 12 t + 24 e^{t} - 24

Schritte zur Lösung

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

Multipliziere aus

(t - u)^{2} e^{u} u^{2} : t^{2} e^{u} u^{2} - 2 t e^{u} u^{3} + e^{u} u^{4}

= \int_{0}^{t} t^{2} e^{u} u^{2} - 2 t e^{u} u^{3} + e^{u} u^{4} d u

Wende die Summenregel an:

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

= \int_{0}^{t} t^{2} e^{u} u^{2} d u - \int_{0}^{t} 2 t e^{u} u^{3} d u + \int_{0}^{t} e^{u} u^{4} d u

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

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

\int_{0}^{t} e^{u} u^{4} d u = t^{4} e^{t} - 4 (t^{3} e^{t} - 3 (t^{2} e^{t} - 2 (e^{t} t - e^{t}))) - 24

= t^{2} (t^{2} e^{t} - 2 (e^{t} t - e^{t}) - 2) - 2 t (t^{3} e^{t} - 3 (t^{2} e^{t} - 2 (t e^{t} - e^{t})) + 6) + t^{4} e^{t} - 4 (t^{3} e^{t} - 3 (t^{2} e^{t} - 2 (e^{t} t - e^{t}))) - 24

Vereinfache

t^{2} (t^{2} e^{t} - 2 (e^{t} t - e^{t}) - 2) - 2 t (t^{3} e^{t} - 3 (t^{2} e^{t} - 2 (t e^{t} - e^{t})) + 6) + t^{4} e^{t} - 4 (t^{3} e^{t} - 3 (t^{2} e^{t} - 2 (e^{t} t - e^{t}))) - 24 : 2 t^{2} e^{t} - 2 t^{2} - 12 t e^{t} - 12 t + 24 e^{t} - 24

= 2 t^{2} e^{t} - 2 t^{2} - 12 t e^{t} - 12 t + 24 e^{t} - 24

\int_{0}^{4} e^{y^{2}} (2 - \frac{y}{2}) d y

\int_{1}^{\infty} e^{- s x} d x

\int_{π}^{3} csc^{2} (θ) d θ

\int_{3}^{\infty} \frac{4}{x ^{3}} d x

\int_{\frac{π}{4}}^{x} cos (2 t) d t