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

Lösung

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

\frac{4}{3}

Dezimale

1.33333 \dots

Schritte zur Lösung

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

Multipliziere aus

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

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

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

\int_{0}^{2} 2 y^{2} d y = \frac{16}{3}

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

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

= \frac{16}{3} - 2 - 2

Vereinfache

\frac{16}{3} - 2 - 2 : \frac{4}{3}

= \frac{4}{3}

- 6 y^{^{''}} + 24 y = 0, y (- 1) = 1, y^{^{'}} (1) = 1

d er i v a t i v e f (x) = \frac{x - 3}{2}

\frac{\partial}{\partial y} (\frac{x}{y ^{2} + x ^{2}})

t a y l or x^{2}, 1

y^{^{''''}} + y^{^{''}} = 0