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_{1}^{5} (\frac{4}{x} - e^{- x}) d x

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