integral from 0 to t^4 of sqrt(u^7)

Lösung

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

\frac{2 t ^{18}}{9}

Schritte zur Lösung

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

u^{7} = u^{\frac{7}{2}},

angenommen

u \geq 0

= \int_{0}^{t^{4}} u^{\frac{7}{2}} d u

Wende die Potenzregel an

= [\frac{2}{9} u^{\frac{9}{2}}]_{0}^{t^{4}}

Berechne die Grenzen:

\frac{2}{9} t^{18}

= \frac{2}{9} t^{18}

Vereinfache

= \frac{2 t ^{18}}{9}

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

\int_{0}^{1} k \cdot (1 - y) y^{n} d y

\int_{0}^{e} (ln (x))^{2} d x

\int_{10}^{20} 3 x^{2} + 6 x + 300 d x

\int_{0}^{1} \frac{3}{8} x^{2} d x