fläche y=e,y=e^x,y=e^{-x}

Lösung

fläche

y = e, y = e^{x}, y = e^{- x}

2

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

f_{i} (x) = f_{j} (x)

e = e^{x} : x = 1

e = e^{- x} : x = - 1

e^{x} = e^{- x} : x = 0

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

= \int_{- 1}^{0} e - e^{- x} d x + \int_{0}^{1} ∣ e^{x} - e ∣ d x

\int_{- 1}^{0} e - e^{- x} d x = 1

\int_{0}^{1} ∣ e^{x} - e ∣ d x = 1

= 1 + 1

Vereinfache

= 2

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 14

x \to 2 + lim (\frac{- 1}{x ^{2} - 4})

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

\int (e^{x}) d x

\frac{d}{d y} (tan (y))