integral from 0 to 1 of ln(y+1)

Lösung

\int_{0}^{1} ln (y + 1) d y

2 ln (2) - 1

Dezimale Notation

0.38629 \dots

Schritte zur Lösung

\int_{0}^{1} ln (y + 1) d y

Wende U-Substitution an

: \int_{1}^{2} ln (u) d u

= \int_{1}^{2} ln (u) d u

Wende die partielle Integration an

: [u ln (u) - \int 1 d u]_{1}^{2}

= [u ln (u) - \int 1 d u]_{1}^{2}

\int 1 d u = u

= [u ln (u) - u]_{1}^{2}

Berechne die Grenzen:

2 ln (2) - 1

= 2 ln (2) - 1

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

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\int_{- 1}^{1} 4 1 - x^{2} d x

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\int_{a}^{b} x^{2} \frac{1}{b - a} d x