integral from 0 to 1 of x^3sqrt(1-x^4)

Lösung

\int_{0}^{1} x^{3} 1 - x^{4} d x

\frac{1}{6}

Dezimale

0.16666 \dots

Schritte zur Lösung

\int_{0}^{1} x^{3} 1 - x^{4} d x

Wende U-Substitution an

= \int_{1}^{0} - \frac{u}{4} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - \frac{u}{4} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{4} \cdot \int_{0}^{1} u d u)

Wende die Potenzregel an

= - (- \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1})

Vereinfache

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{1}

Berechne die Grenzen:

\frac{2}{3}

= \frac{1}{4} \cdot \frac{2}{3}

Vereinfache

= \frac{1}{6}

\frac{\partial}{\partial x} (8 y x)

h \to 0 lim (\frac{( 2 + h ) - 2}{h})

d er i v a t i v e cos (2 x + 1)

d er i v a t i v e y = \frac{1 + 2 x}{3 - 4 x}

x \to 0 + lim (\frac{e ^{x}}{1 - e ^{x}})