integral from 0 to 3 of (3000)/(1+0.25t)

Lösung

\int_{0}^{3} \frac{3000}{1 + 0.25 t} d t

6715.38945 \dots

Schritte zur Lösung

\int_{0}^{3} \frac{3000}{1 + 0.25 t} d t

Entferne die Konstante:

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

= 3000 \cdot \int_{0}^{3} \frac{1}{1 + 0.25 t} d t

Wende U-Substitution an

= 3000 \cdot \int_{1}^{1.75} \frac{1}{0.25 u} d u

Entferne die Konstante:

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

= 3000 \cdot \frac{1}{0.25} \cdot \int_{1}^{1.75} \frac{1}{u} d u

Vereinfache

= 3000 \cdot 4 \cdot \int_{1}^{1.75} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3000 \cdot 4 [ln ∣ u ∣]_{1}^{1.75}

Vereinfache

= 12000 [ln ∣ u ∣]_{1}^{1.75}

Berechne die Grenzen:

0.55961 \dots

= 12000 \cdot 0.55961 \dots

Vereinfache

= 6715.38945 \dots

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