integral from 0 to 1 of 18 1/(x^p)

Lösung

\int_{0}^{1} 18 \frac{1}{x ^{p}} d x

\frac{18}{- p + 1}

Schritte zur Lösung

\int_{0}^{1} 18 \cdot \frac{1}{x ^{p}} d x

Entferne die Konstante:

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

= 18 \cdot \int_{0}^{1} \frac{1}{x ^{p}} d x

Wende die Potenzregel an

= 18 [\frac{x ^{- p + 1}}{- p + 1}]_{0}^{1}

Berechne die Grenzen:

\frac{1}{- p + 1}

= 18 \cdot \frac{1}{- p + 1}

Vereinfache

= \frac{18}{- p + 1}

t an g e n t y = x (1 - 2 x)^{3}, (1, - 1)

\frac{\partial}{\partial y} (x^{3} ln (y^{6}))

n \to \infty lim (n^{2} + 2)

\int (\frac{x + 1}{4 x}) d x

y^{2} d y + x^{2} d x = 0