integral from 2 to 5 of t^3ln(4t)

Lösung

\int_{2}^{5} t^{3} ln (4 t) d t

- 12 ln (2) - \frac{609}{16} + \frac{625}{4} ln (20)

Dezimale

421.70290 \dots

Schritte zur Lösung

\int_{2}^{5} t^{3} ln (4 t) d t

Wende die partielle Integration an

= [\frac{1}{4} t^{4} ln (4 t) - \int \frac{t ^{3}}{4} d t]_{2}^{5}

\int \frac{t ^{3}}{4} d t = \frac{t ^{4}}{16}

= [\frac{1}{4} t^{4} ln (4 t) - \frac{t ^{4}}{16}]_{2}^{5}

Berechne die Grenzen:

- 12 ln (2) - \frac{609}{16} + \frac{625}{4} ln (20)

= - 12 ln (2) - \frac{609}{16} + \frac{625}{4} ln (20)

\int_{0}^{2} x sin (nπ x) d x

\int_{- 2}^{1} (2 - y) - y^{2} d y

\int_{0.1}^{0.25} \frac{x ^{3} + 1}{6} d x

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

\int_{0}^{4} (\frac{4}{x}) - (4 x) d x