integral from-1 to 1 of x+y+z

Lösung

\int_{- 1}^{1} x + y + z d x

2 y + 2 z

Schritte zur Lösung

\int_{- 1}^{1} x + y + z 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_{- 1}^{1} y d x + \int_{- 1}^{1} z d x + \int_{- 1}^{1} x d x

\int_{- 1}^{1} y d x = 2 y

\int_{- 1}^{1} z d x = 2 z

\int_{- 1}^{1} x d x = 0

= 2 y + 2 z + 0

Vereinfache

= 2 y + 2 z

\int_{0}^{r} 0.5 e^{- 0.5 x} d x

\int_{1}^{2} x^{6} ln (x) d x

\int_{0}^{π} cos (x) - 0.5 d x

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

\int_{- 3}^{9} - x cos (x) d x