integral from 1 to 3 of 3/(7x-2)

Lösung

\int_{1}^{3} \frac{3}{7 x - 2} d x

\frac{3}{7} (ln (19) - ln (5))

Dezimale

0.57214 \dots

Schritte zur Lösung

\int_{1}^{3} \frac{3}{7 x - 2} d x

Entferne die Konstante:

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

= 3 \cdot \int_{1}^{3} \frac{1}{7 x - 2} d x

Wende U-Substitution an

= 3 \cdot \int_{5}^{19} \frac{1}{7 u} d u

Entferne die Konstante:

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

= 3 \cdot \frac{1}{7} \cdot \int_{5}^{19} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{7} [ln ∣ u ∣]_{5}^{19}

Vereinfache

3 \cdot \frac{1}{7} [ln ∣ u ∣]_{5}^{19} : \frac{3}{7} [ln ∣ u ∣]_{5}^{19}

= \frac{3}{7} [ln ∣ u ∣]_{5}^{19}

Berechne die Grenzen:

ln (19) - ln (5)

= \frac{3}{7} (ln (19) - ln (5))

\int_{x}^{4} \frac{6 ( y - x )}{64} d y

\int_{0}^{l o g_{6} (2)} 6^{x} d x

\int_{- 1}^{1} \frac{x}{1 + x ^{4}} d x

\int_{0}^{1} 1.5 (1 - x^{2}) x d x

\int_{0}^{π} sin^{2} (y) d y