integral from 0 to 2 of-2t+2

Lösung

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

0

Schritte zur Lösung

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

\int_{0}^{2} 2 t d t = 4

\int_{0}^{2} 2 d t = 4

= - 4 + 4

Vereinfache

= 0

\int_{0}^{2} 8^{x} d x

\int_{0}^{2} \frac{40}{x ^{2} + 4} d x

\int_{9}^{x^{3}} \frac{1}{p ^{2}} d p

\int_{0}^{4} t^{3} d t

\int_{- 2}^{4} \frac{y + 4}{2} - \frac{y ^{2}}{4} d y