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

Lösung

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

ln (2)

Dezimale

0.69314 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{2}^{4})

Vereinfache

= [ln ∣ u ∣]_{2}^{4}

Berechne die Grenzen:

ln (2)

= ln (2)

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

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

\int_{0}^{1} ∣ x (2 x - 1) ∣ d x

\int_{0}^{3} e^{- t^{2}} d t

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