integral from-1 to 0 of (x+1)^3

Lösung

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

\frac{1}{4}

Dezimale

0.25

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{0}^{1} u^{3} d u

Wende die Potenzregel an

= [\frac{u ^{4}}{4}]_{0}^{1}

Berechne die Grenzen:

\frac{1}{4}

= \frac{1}{4}

\frac{d}{d x} (7 x - 4)

(y^{3} - 1) e x d x + 3 y^{2} (e x + 1) d y = 0

x \to 0 lim (4 x cot (x))

\int_{0}^{1} 8 e^{2 x} d x

d er i v a t i v e 2 e^{x} + \frac{8}{3 x}