integral from-1 to 1 of (u^7+1)(3u^2+1)

Lösung

\int_{- 1}^{1} (u^{7} + 1) (3 u^{2} + 1) d u

4

Schritte zur Lösung

\int_{- 1}^{1} (u^{7} + 1) (3 u^{2} + 1) d u

Multipliziere aus

(u^{7} + 1) (3 u^{2} + 1) : 3 u^{9} + u^{7} + 3 u^{2} + 1

= \int_{- 1}^{1} 3 u^{9} + u^{7} + 3 u^{2} + 1 d u

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} 3 u^{9} d u + \int_{- 1}^{1} u^{7} d u + \int_{- 1}^{1} 3 u^{2} d u + \int_{- 1}^{1} 1 d u

\int_{- 1}^{1} 3 u^{9} d u = 0

\int_{- 1}^{1} u^{7} d u = 0

\int_{- 1}^{1} 3 u^{2} d u = 2

\int_{- 1}^{1} 1 d u = 2

= 0 + 0 + 2 + 2

Vereinfache

= 4

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