integral from 0 to 2pi of-3sin^2(t)+9t

Lösung

\int_{0}^{2 π} - 3 sin^{2} (t) + 9 t d t

- 3 π + 18 π^{2}

Dezimale

168.22810 \dots

Schritte zur Lösung

\int_{0}^{2 π} - 3 sin^{2} (t) + 9 t 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}^{2 π} 3 sin^{2} (t) d t + \int_{0}^{2 π} 9 t d t

\int_{0}^{2 π} 3 sin^{2} (t) d t = 3 π

\int_{0}^{2 π} 9 t d t = 18 π^{2}

= - 3 π + 18 π^{2}

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