integral from 2 to 5 of (t^2+2t)

Lösung

\int_{2}^{5} (t^{2} + 2 t) d t

60

Schritte zur Lösung

\int_{2}^{5} t^{2} + 2 t 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_{2}^{5} t^{2} d t + \int_{2}^{5} 2 t d t

\int_{2}^{5} t^{2} d t = 39

\int_{2}^{5} 2 t d t = 21

= 39 + 21

Vereinfache

= 60

\int_{0}^{2} y e^{- y^{2}} d y

\int_{- 1}^{2} ∣ x - 1 ∣ sin (x) d x

\int_{0}^{2} sin (x^{2}) x d x

\int_{0}^{- 1} x^{3} 1 - x^{2} d x

\int_{0}^{1} 2 x arccos (x) d x