integral from-1 to 0 of (1+x)

Lösung

\int_{- 1}^{0} (1 + x) d x

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

\int_{- 1}^{0} 1 + 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_{- 1}^{0} 1 d x + \int_{- 1}^{0} x d x

\int_{- 1}^{0} 1 d x = 1

\int_{- 1}^{0} x d x = - \frac{1}{2}

= 1 - \frac{1}{2}

Vereinfache

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

= \frac{1}{2}

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