integral from 0 to 2 of 1/(x+4)

Lösung

\int_{0}^{2} \frac{1}{x + 4} d x

ln (\frac{3}{2})

Dezimale

0.40546 \dots

Schritte zur Lösung

\int_{0}^{2} \frac{1}{x + 4} d x

Wende U-Substitution an

= \int_{4}^{6} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{4}^{6}

Berechne die Grenzen:

ln (6) - 2 ln (2)

= ln (6) - 2 ln (2)

Vereinfache

= ln (\frac{3}{2})

\int_{1}^{6} t^{2} - 2 t - 8 d t

\int_{1}^{- 1} x^{2} d x

\int_{x^{2}}^{x} (x^{2} + y^{2}) d y

\int_{- 1}^{1} x^{7} d x

\int_{0}^{1} x (x^{2} + 6) d x