integral from 2 to 4 of 1/(3t+4)

Lösung

\int_{2}^{4} \frac{1}{3 t + 4} d t

\frac{1}{3} ln (\frac{8}{5})

Dezimale

0.15666 \dots

Schritte zur Lösung

\int_{2}^{4} \frac{1}{3 t + 4} d t

Wende U-Substitution an

= \int_{10}^{16} \frac{1}{3 u} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{10}^{16} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{10}^{16}

Berechne die Grenzen:

4 ln (2) - ln (10)

= \frac{1}{3} (4 ln (2) - ln (10))

Vereinfache

= \frac{1}{3} ln (\frac{8}{5})

\int_{0}^{\frac{1}{2}} (x e^{3 x})^{2} d x

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