fläche e^x,e^{-2x},ln(4)

Lösung

fläche

e^{x}, e^{- 2 x}, ln (4)

0.09977 \dots

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

e^{x} = e^{- 2 x} : x = 0

e^{x} = ln (4) : x = ln (2 ln (2))

e^{- 2 x} = ln (4) : x = - \frac{ln ( 2 ln ( 2 ) )}{2}

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

= \int_{- \frac{l n ( 2 l n ( 2 ) )}{2}}^{0} e^{- 2 x} - ln (4) d x + \int_{0}^{l n (2 l n (2))} ∣ e^{x} - ln (4) ∣ d x

\int_{- \frac{l n ( 2 l n ( 2 ) )}{2}}^{0} e^{- 2 x} - ln (4) d x = 0.03325 \dots

\int_{0}^{l n (2 l n (2))} ∣ e^{x} - ln (4) ∣ d x = 0.06651 \dots

= 0.03325 \dots + 0.06651 \dots

Vereinfache

= 0.09977 \dots

\int_{0}^{3} x^{3} - x^{2} - 6 x d x

\frac{\partial}{\partial x} ((x + 20) y)

\int (x + 5)^{4} d x

\frac{d}{d x} (ln (2 sin (x)))

d er i v a t i v e y = (3 - e^{x})^{3}