integral from y to 1 of y(e^x-1)

Lösung

\int_{y}^{1} y (e^{x} - 1) d x

y (- 1 + y + e - e^{y})

Schritte zur Lösung

\int_{y}^{1} y (e^{x} - 1) d x

Entferne die Konstante:

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

= y \cdot \int_{y}^{1} - 1 + e^{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

= y (- \int_{y}^{1} 1 d x + \int_{y}^{1} e^{x} d x)

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

\int_{y}^{1} e^{x} d x = e - e^{y}

= y (- (1 - y) + e - e^{y})

Vereinfache

= y (- 1 + y + e - e^{y})

\int_{0}^{2} \frac{1}{3 x + 4} d x

\int_{0}^{1} \frac{2}{x ^{2} - 4} d x

\int_{\frac{π}{2}}^{π} sin (θ) d θ

\int_{1}^{4} \frac{e ^{x}}{x} d x

\int_{0}^{\infty} e^{- x} \cdot x d x