integral from 1 to e^4 of sqrt(x)ln(x)

Lösung

\int_{1}^{e^{4}} x ln (x) d x

\frac{20 e ^{6} + 4}{9}

Dezimale

896.95287 \dots

Schritte zur Lösung

\int_{1}^{e^{4}} x ln (x) d x

Wende die partielle Integration an

= [\frac{1}{3} (2 x^{\frac{3}{2}} ln (x) - 3 \cdot \int \frac{2 x ^{\frac{1}{2}}}{3} d x)]_{1}^{e^{4}}

\int \frac{2 x ^{\frac{1}{2}}}{3} d x = \frac{4}{9} x^{\frac{3}{2}}

= [\frac{1}{3} (2 x^{\frac{3}{2}} ln (x) - 3 \cdot \frac{4}{9} x^{\frac{3}{2}})]_{1}^{e^{4}}

3 \cdot \frac{4}{9} x^{\frac{3}{2}} = \frac{4}{3} x^{\frac{3}{2}}

= [\frac{1}{3} (2 x^{\frac{3}{2}} ln (x) - \frac{4}{3} x^{\frac{3}{2}})]_{1}^{e^{4}}

Berechne die Grenzen:

\frac{20 e ^{6}}{9} + \frac{4}{9}

= \frac{20 e ^{6}}{9} + \frac{4}{9}

Vereinfache

= \frac{20 e ^{6} + 4}{9}

\int_{2}^{4} 9 x^{2} d x

\int_{0}^{\frac{π}{4}} 3 sin (x) d x

\int_{- 1}^{1} arcsin (x) d x

\int_{- 3}^{3} (x^{3}) d x

\int_{- 8}^{0} (2 + 64 - x^{2}) d x