integral from 0 to 1 of (1-x)^3

Lösung

\int_{0}^{1} (1 - x)^{3} d x

\frac{1}{4}

Dezimale

0.25

Schritte zur Lösung

\int_{0}^{1} (1 - x)^{3} d x

Multipliziere aus

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

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

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

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

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

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

= 1 - \frac{3}{2} + 1 - \frac{1}{4}

Vereinfache

1 - \frac{3}{2} + 1 - \frac{1}{4} : \frac{1}{4}

= \frac{1}{4}

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

\int_{0}^{2} (x^{2} - 3 x + 2) d x

\int_{- 1}^{1} y d y

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

\int_{0}^{π} sin (x) sin (2 x) d x