integral from 0 to 1 of 1/(3x+1)

Lösung

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

\frac{2}{3} ln (2)

Dezimale

0.46209 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{4} \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_{1}^{4} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{1}^{4}

Berechne die Grenzen:

2 ln (2)

= \frac{1}{3} \cdot 2 ln (2)

Vereinfache

= \frac{2}{3} ln (2)

\int_{1}^{3} 3 x^{3} d x

\int_{1}^{2} 4 x (x^{2} + 3) d x

\int_{0}^{3} π (x)^{2} d x

\int_{- \frac{π}{2}}^{π} 4 cos (x) d x

\int_{- 3}^{0} x x + 3 d x