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

Lösung

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

\frac{1}{4}

Dezimale

0.25

Schritte zur Lösung

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

Multipliziere aus

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

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

\int_{0}^{1} x^{3} d x = \frac{1}{4}

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

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

= \frac{1}{4} - 1 + 1

Vereinfache

= \frac{1}{4}

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

\int_{0}^{3} 3^{3 x} d x

\int_{2}^{5} π (x^{2} - \frac{1}{5} x^{4}) d x

\int_{0}^{\frac{1}{2}} arcsin^{2} (x) d x

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