integral from 1 to e of yln(y)

Lösung

\int_{1}^{e} y ln (y) d y

\frac{e ^{2} + 1}{4}

Dezimale

2.09726 \dots

Schritte zur Lösung

\int_{1}^{e} y ln (y) d y

Wende die partielle Integration an

= [\frac{1}{2} (y^{2} ln (y) - 2 \cdot \int \frac{y}{2} d y)]_{1}^{e}

\int \frac{y}{2} d y = \frac{y ^{2}}{4}

= [\frac{1}{2} (y^{2} ln (y) - 2 \cdot \frac{y ^{2}}{4})]_{1}^{e}

2 \cdot \frac{y ^{2}}{4} = \frac{y ^{2}}{2}

= [\frac{1}{2} (y^{2} ln (y) - \frac{y ^{2}}{2})]_{1}^{e}

Berechne die Grenzen:

\frac{e ^{2}}{4} + \frac{1}{4}

= \frac{e ^{2}}{4} + \frac{1}{4}

Vereinfache

= \frac{e ^{2} + 1}{4}

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

\int_{0}^{0} f (t) d t

\int_{0}^{\infty} cos (2 x) d x

\int_{2}^{23} \frac{1}{x + 2} d x

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