integral from 0 to 1 of (2t+1)

Lösung

\int_{0}^{1} (2 t + 1) d t

2

Schritte zur Lösung

\int_{0}^{1} 2 t + 1 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}^{1} 2 t d t + \int_{0}^{1} 1 d t

\int_{0}^{1} 2 t d t = 1

\int_{0}^{1} 1 d t = 1

= 1 + 1

Vereinfache

= 2

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