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

Lösung

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

ln (2)

Dezimale

0.69314 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{2}^{4} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{4}

Berechne die Grenzen:

ln (2)

= ln (2)

x \to 0 lim (\frac{( cos ( 11 x ) - cos ( 6 x ) )}{x})

d er i v a t i v e f (x) = x^{3} sin (x)

\frac{d}{d x} (\frac{x ^{2} - 6 x + 1}{( x ^{2} - 1 ) ^{2}})

\int cos (x) (5 + 5 sin^{2} (x)) d x

\int_{0}^{1} (3 1 + 7 x) d x