integral from 0 to 4 of (x-2)(2x+1)

Lösung

\int_{0}^{4} (x - 2) (2 x + 1) d x

\frac{32}{3}

Dezimale

10.66666 \dots

Schritte zur Lösung

\int_{0}^{4} (x - 2) (2 x + 1) d x

Multipliziere aus

(x - 2) (2 x + 1) : 2 x^{2} - 3 x - 2

= \int_{0}^{4} 2 x^{2} - 3 x - 2 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_{0}^{4} 2 x^{2} d x - \int_{0}^{4} 3 x d x - \int_{0}^{4} 2 d x

\int_{0}^{4} 2 x^{2} d x = \frac{128}{3}

\int_{0}^{4} 3 x d x = 24

\int_{0}^{4} 2 d x = 8

= \frac{128}{3} - 24 - 8

Vereinfache

\frac{128}{3} - 24 - 8 : \frac{32}{3}

= \frac{32}{3}

\int_{0}^{1} e^{x^{\frac{1}{2}}} d x

\int_{- 2}^{4} - x + 4 d x

\int_{- \frac{π}{4}}^{\frac{π}{4}} cos^{2} (x) d x

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