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

Lösung

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

ln (7) - ln (3)

Dezimale

0.84729 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{3}^{7}

Berechne die Grenzen:

ln (7) - ln (3)

= ln (7) - ln (3)

\int_{0}^{1} - 6 x^{3} (3 x^{4} - 2)^{5} d x

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

\int_{0}^{5} x^{2} - 9 d x

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

\int_{0}^{\frac{π}{4}} (9 sec^{2} (x)) d x