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

Lösung

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

\frac{1}{3}

Dezimale

0.33333 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Multipliziere aus

y (1 - y) : y - y^{2}

= 2 \cdot \int_{0}^{1} y - y^{2} d y

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

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

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

= 2 (\frac{1}{2} - \frac{1}{3})

Vereinfache

2 (\frac{1}{2} - \frac{1}{3}) : \frac{1}{3}

= \frac{1}{3}

\int_{0}^{5} 2 e^{0.2 t} - t d t

\int_{1}^{2} (3 x^{5} - 2 x^{3}) d x

\int_{0}^{t} e^{a t} d t

\int_{- 1}^{5} 2 x + 6 d x

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