integral from 0 to 1 of 2/3 (x+2y)

Lösung

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

\frac{1}{3} + \frac{4}{3} y

Schritte zur Lösung

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

Entferne die Konstante:

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

= \frac{2}{3} \cdot \int_{0}^{1} x + 2 y 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

= \frac{2}{3} (\int_{0}^{1} x d x + \int_{0}^{1} 2 y d x)

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

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

= \frac{2}{3} (\frac{1}{2} + 2 y)

Vereinfache

\frac{2}{3} (\frac{1}{2} + 2 y) : \frac{1}{3} + \frac{4}{3} y

= \frac{1}{3} + \frac{4}{3} y

\frac{d}{d x} (a x + cot (a x))

x \to - 4 + lim (\frac{- 4 x}{x + 4})

f (x) = 2 sin^{2} (x)

\frac{\partial}{\partial x} (7 y e^{x})

\frac{d y}{d x} = \frac{2 x + sec ^{2} ( x )}{2 y}, y (0) = - 7