integral from 1 to 4 of (5-2t)

Lösung

\int_{1}^{4} (5 - 2 t) d t

0

Schritte zur Lösung

\int_{1}^{4} 5 - 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_{1}^{4} 5 d t - \int_{1}^{4} 2 t d t

\int_{1}^{4} 5 d t = 15

\int_{1}^{4} 2 t d t = 15

= 15 - 15

Vereinfache

= 0

\int_{1}^{e} 1 x ln (x) d x

\int_{0}^{1} x (4 - x) d x

\int_{- 1}^{2} x^{2} + x + 1 d x

\int_{0}^{1} t cosh (t) d t

\int_{0}^{t^{4}} u^{7} d u