integral from 0 to 1 of 2x^3-6x^2+4x

Lösung

\int_{0}^{1} 2 x^{3} - 6 x^{2} + 4 x d x

\frac{1}{2}

Dezimale

0.5

Schritte zur Lösung

\int_{0}^{1} 2 x^{3} - 6 x^{2} + 4 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_{0}^{1} 2 x^{3} d x - \int_{0}^{1} 6 x^{2} d x + \int_{0}^{1} 4 x d x

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

\int_{0}^{1} 6 x^{2} d x = 2

\int_{0}^{1} 4 x d x = 2

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

Vereinfache

= \frac{1}{2}

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