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Beliebt Rechnen >

integral from 1 to e^2 of x^2ln(x)

Lösung

\int_{1}^{e^{2}} x^{2} ln (x) d x

Lösung

\frac{5 e ^{6} + 1}{9}

+1

Dezimale

224.23821 \dots

Schritte zur Lösung

\int_{1}^{e^{2}} x^{2} ln (x) d x

Wende die partielle Integration an

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

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

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

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

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

Berechne die Grenzen:

\frac{5 e ^{6}}{9} + \frac{1}{9}

= \frac{5 e ^{6}}{9} + \frac{1}{9}

Vereinfache

= \frac{5 e ^{6} + 1}{9}

Graph

Beliebte Beispiele

derivative (x+4)/(x^3+x-5)

d er i v a t i v e \frac{x + 4}{x ^{3} + x - 5}

fläche y=2x,x=5,y= 2/x

a re a y = 2 x, x = 5, y = \frac{2}{x}

integral of (x+1)/(x^3+x^2-6x)

\int \frac{x + 1}{x ^{3} + x ^{2} - 6 x} d x

tangent y=x^2,\at x=-0.5

t an g e n t y = x^{2}, at x = - 0.5

derivative of 1+x^2y^2

\frac{d}{d x} (1 + x^{2} y^{2})