fläche e^x,e^{-4x},ln(3)

Lösung

fläche

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

0.00588 \dots

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

e^{x} = e^{- 4 x} : x = 0

e^{x} = ln (3) : x = ln (ln (3))

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

= \int_{- \frac{l n ( l n ( 3 ) )}{4}}^{0} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{0}^{l n (l n (3))} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- \frac{l n ( l n ( 3 ) )}{4}}^{0} e^{- 4 x} - ln (3) d x + \int_{0}^{l n (l n (3))} ∣ e^{x} - ln (3) ∣ d x

\int_{- \frac{l n ( l n ( 3 ) )}{4}}^{0} e^{- 4 x} - ln (3) d x = 0.00117 \dots

\int_{0}^{l n (l n (3))} ∣ e^{x} - ln (3) ∣ d x = 0.00470 \dots

= 0.00117 \dots + 0.00470 \dots

Vereinfache

= 0.00588 \dots

d er i v a t i v e f (x) = 7 e^{x} + x

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\frac{d}{d x} (lo g_{10} (i) cos (x))