integral from 1 to 2 of 1/(3-z)

Lösung

\int_{1}^{2} \frac{1}{3 - z} d z

ln (2)

Dezimale

0.69314 \dots

Schritte zur Lösung

\int_{1}^{2} \frac{1}{3 - z} d z

Wende U-Substitution an

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

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

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- [ln ∣ u ∣]_{1}^{2})

Vereinfache

= [ln ∣ u ∣]_{1}^{2}

Berechne die Grenzen:

ln (2)

= ln (2)

\int (\frac{x ^{7} - 4 x ^{5} + 11}{x ^{5}}) d x

\frac{\partial}{\partial x} (e^{x} y x + y e^{x} + y^{2} e x^{x})

x \to 1 lim ((1 - 2 x)^{\frac{1}{x}})

d er i v a t i v e f (x) = \frac{t ^{3} + 3 t}{t ^{3 - 4 t + 3}}

\frac{\partial}{\partial x} (3 sec^{2} (2 x + 3 y))