integral from 0 to pi of (2+sin(x))

Lösung

\int_{0}^{π} (2 + sin (x)) d x

2 π + 2

Dezimale

8.28318 \dots

Schritte zur Lösung

\int_{0}^{π} 2 + sin (x) 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_{0}^{π} 2 d x + \int_{0}^{π} sin (x) d x

\int_{0}^{π} 2 d x = 2 π

\int_{0}^{π} sin (x) d x = 2

= 2 π + 2

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