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

Lösung

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

\frac{1}{10}

Dezimale

0.1

Schritte zur Lösung

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Multipliziere aus

x (- x + 1)^{3} : - x^{4} + 3 x^{3} - 3 x^{2} + x

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

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

\int_{0}^{1} x^{4} d x = \frac{1}{5}

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

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

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

= 2 (- \frac{1}{5} + \frac{3}{4} - 1 + \frac{1}{2})

Vereinfache

2 (- \frac{1}{5} + \frac{3}{4} - 1 + \frac{1}{2}) : \frac{1}{10}

= \frac{1}{10}

\int_{3}^{\infty} e^{- s t} t d t

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

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

\int_{a}^{\infty} \frac{1}{x ^{3}} d x

\int_{0}^{r} r^{2} - r^{2} r d r