integral from 0 to 1 of t^2+t

Lösung

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

\frac{5}{6}

Dezimale

0.83333 \dots

Schritte zur Lösung

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

\int_{0}^{1} t^{2} d t = \frac{1}{3}

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

= \frac{1}{3} + \frac{1}{2}

Vereinfache

\frac{1}{3} + \frac{1}{2} : \frac{5}{6}

= \frac{5}{6}

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