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

Lösung

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

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

Dezimale

0.20273 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{4}^{6} \frac{1}{2 u} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{4}^{6} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{4}^{6}

Berechne die Grenzen:

ln (6) - 2 ln (2)

= \frac{1}{2} (ln (6) - 2 ln (2))

Vereinfache

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

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

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\int_{- \frac{π}{4}}^{0} sec (x) tan (x) d x