integral from 0 to 2 of 1/((3x+4))

Lösung

\int_{0}^{2} \frac{1}{( 3 x + 4 )} d x

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

Dezimale

0.30543 \dots

Schritte zur Lösung

\int_{0}^{2} \frac{1}{3 x + 4} d x

Wende U-Substitution an

= \int_{4}^{10} \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_{4}^{10} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{4}^{10}

Berechne die Grenzen:

ln (10) - 2 ln (2)

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

Vereinfache

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

\int_{0}^{\frac{π}{2}} x^{2} cos (4 x) d x

\int_{0}^{5} 3 x + 2 d x

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

\int_{0}^{12} \frac{1}{1 + x ^{2}} d x

\int_{3}^{\infty} \frac{3}{x ^{6}} d x