integral from 2 to 5 of 1/(2-5t)

Lösung

\int_{2}^{5} \frac{1}{2 - 5 t} d t

\frac{1}{5} (3 ln (2) - ln (23))

Dezimale

- 0.21121 \dots

Schritte zur Lösung

\int_{2}^{5} \frac{1}{2 - 5 t} d t

Wende U-Substitution an

= \int_{- 8}^{- 23} - \frac{1}{5 u} d u

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

= - \int_{- 23}^{- 8} - \frac{1}{5 u} d u

Entferne die Konstante:

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

= - (- \frac{1}{5} \cdot \int_{- 23}^{- 8} \frac{1}{u} d u)

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{5} [ln ∣ u ∣]_{- 23}^{- 8})

Vereinfache

= \frac{1}{5} [ln ∣ u ∣]_{- 23}^{- 8}

Berechne die Grenzen:

3 ln (2) - ln (23)

= \frac{1}{5} (3 ln (2) - ln (23))

9 y^{^{''}} - 6 y^{^{'}} + 4 y = 0

\frac{d}{d x} ((1 - x) e^{- x})

x \to 0 lim (\frac{4 sin ( 3 x )}{2 tan ( 5 x )})

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

\int_{0}^{1} \frac{3 x ^{2}}{1 + x ^{3}} d x