fläche e^x,e^{-3x},ln(4)

Lösung

fläche

e^{x}, e^{- 3 x}, ln (4)

0.08868 \dots

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

e^{x} = e^{- 3 x} : x = 0

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

e^{- 3 x} = ln (4) : x = - \frac{ln ( 2 ln ( 2 ) )}{3}

= \int_{- \frac{l n ( 2 l n ( 2 ) )}{3}}^{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 ) )}{3}}^{0} e^{- 3 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 ) )}{3}}^{0} e^{- 3 x} - ln (4) d x = 0.02217 \dots

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

= 0.02217 \dots + 0.06651 \dots

Vereinfache

= 0.08868 \dots

\frac{d x}{d t} = 1 - x^{2}

\frac{d}{d x} (\frac{3 ( x ^{2} + 2 ) ^{3}}{5})

\frac{d x}{d t} = a x

t an g e n t \frac{( x + 1 )}{( x ^{2} + x + 1 )}

\int (2 x^{2} - \frac{1}{3})^{3} d x