integral from 1 to 3 of (x^2+x+4)

Lösung

\int_{1}^{3} (x^{2} + x + 4) d x

\frac{62}{3}

Dezimale

20.66666 \dots

Schritte zur Lösung

\int_{1}^{3} x^{2} + x + 4 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}^{3} x^{2} d x + \int_{1}^{3} x d x + \int_{1}^{3} 4 d x

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} x d x = 4

\int_{1}^{3} 4 d x = 8

= \frac{26}{3} + 4 + 8

Vereinfache

\frac{26}{3} + 4 + 8 : \frac{62}{3}

= \frac{62}{3}

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