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

Lösung

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

\frac{140}{3}

Dezimale

46.66666 \dots

Schritte zur Lösung

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

\int_{0}^{4} x^{3} d x = 64

\int_{0}^{4} x^{2} d x = \frac{64}{3}

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

= 64 - \frac{64}{3} + 4

Vereinfache

64 - \frac{64}{3} + 4 : \frac{140}{3}

= \frac{140}{3}

\int_{1}^{4} - x^{2} + 8 x d x

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