integral from 0 to 5 of 2t+2

Lösung

\int_{0}^{5} 2 t + 2 d t

35

Schritte zur Lösung

\int_{0}^{5} 2 t + 2 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_{0}^{5} 2 t d t + \int_{0}^{5} 2 d t

\int_{0}^{5} 2 t d t = 25

\int_{0}^{5} 2 d t = 10

= 25 + 10

Vereinfache

= 35

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

\int_{0}^{1} e^{- \frac{1}{2} x^{2}} d x

\int_{0}^{1} a x^{2} d x

\int_{0}^{2} \frac{8 x}{2 x ^{2} + 1} d x

\int_{0}^{2} - 4 d x