integral from 1 to e of 1xln(x)

Lösung

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

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

Dezimale

2.09726 \dots

Schritte zur Lösung

\int_{1}^{e} 1 \cdot x ln (x) d x

Vereinfache

= \int_{1}^{e} x ln (x) d x

Wende die partielle Integration an

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

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

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

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

= [\frac{1}{2} (x^{2} ln (x) - \frac{x ^{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}^{1} x (4 - x) d x

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

\int_{0}^{1} t cosh (t) d t

\int_{0}^{t^{4}} u^{7} d u

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