integral from 0 to 12 of 1/(x+12)

Lösung

\int_{0}^{12} \frac{1}{x + 12} d x

ln (2)

Dezimale

0.69314 \dots

Schritte zur Lösung

\int_{0}^{12} \frac{1}{x + 12} d x

Wende U-Substitution an

= \int_{12}^{24} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{12}^{24}

Berechne die Grenzen:

ln (24) - ln (12)

= ln (24) - ln (12)

Vereinfache

= ln (2)