integral from 0 to t of 1/(t+10)

Lösung

\int_{0}^{t} \frac{1}{t + 10} d t

ln ∣ t + 10 ∣ - ln (10)

Schritte zur Lösung

\int_{0}^{t} \frac{1}{t + 10} d t

Wende U-Substitution an

= \int_{10}^{t + 10} \frac{1}{u} d u

Nutze das gemeinsame Integral :

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{10}^{t + 10}

Berechne die Grenzen:

ln ∣ t + 10 ∣ - ln (10)

= ln ∣ t + 10 ∣ - ln (10)

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

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

\int_{- 1}^{3} \frac{1}{( y + 2 ) ^{3}} d y

\int_{3}^{4} 2^{- x} d x

\int_{0}^{l} (\frac{x}{l})^{2} d x