integral from 1 to 4 of 1sqrt(t)ln(t)

Lösung

\int_{1}^{4} 1 t ln (t) d t

- \frac{28}{9} + \frac{32}{3} ln (2)

Dezimale

4.28245 \dots

Schritte zur Lösung

\int_{1}^{4} 1 \cdot t ln (t) d t

Vereinfache

= \int_{1}^{4} t ln (t) d t

Wende die partielle Integration an

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

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

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

Berechne die Grenzen:

- \frac{28}{9} + \frac{32}{3} ln (2)

= - \frac{28}{9} + \frac{32}{3} ln (2)

\frac{d x}{d t} = \frac{1}{x e ^{t + 3 x}}

a re a 3 x^{2} + 8 x + 2, 2 x^{2} + 4 x - 1

a re a x^{2}, x + 2, x = - 2

\int \frac{5}{x ^{8}} d x

t a y l or sin^{2} (x), 0