integral from 0 to 5 of (2x+5)

Lösung

\int_{0}^{5} (2 x + 5) d x

50

Schritte zur Lösung

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

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 x d x + \int_{0}^{5} 5 d x

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

\int_{0}^{5} 5 d x = 25

= 25 + 25

Vereinfache

= 50

\int_{2}^{4} e^{- s t} (t^{2}) d t

\int_{1}^{3} (1 + \frac{1}{u} + \frac{1}{u ^{2}}) d u

\int_{0}^{2 π} sin^{2} (n x) d x

\int_{0}^{x} \frac{3}{4} x (2 - x) d x

\int_{0}^{7} - x^{2} + 7 x d x