integral from 0 to 1 of (t-1)

Lösung

\int_{0}^{1} (t - 1) d t

- \frac{1}{2}

Dezimale

- 0.5

Schritte zur Lösung

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

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

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

= \frac{1}{2} - 1

Vereinfache

\frac{1}{2} - 1 : - \frac{1}{2}

= - \frac{1}{2}

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