integral from-1 to 8 of 1/(\sqrt[3]{x)}

Lösung

\int_{- 1}^{8} \frac{1}{3 x} d x

\frac{9}{2}

Dezimale

4.5

Schritte zur Lösung

\int_{- 1}^{8} \frac{1}{3 x} d x

Bestimme unbestimmte (Singularitäts-)Punkte:

x = 0

Wenn

b, a < b < c, f (b) =

unbestimmt

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{- 1}^{0} \frac{1}{3 x} d x + \int_{0}^{8} \frac{1}{3 x} d x

\int_{- 1}^{0} \frac{1}{3 x} d x = - \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2}

\int_{0}^{8} \frac{1}{3 x} d x = 6

= - \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2} + 6

Vereinfache

- \frac{3 ( - 1 ) ^{\frac{2}{3}}}{2} + 6 : \frac{9}{2}

= \frac{9}{2}

\int \frac{3}{x ^{2} + x - 2} d x

\int \frac{arccos ( x )}{1 - x ^{2}} d x

\int t (t^{3} - 1) d t

\frac{d}{d x} (\frac{e ^{x} + cos ( x )}{x ^{2}})

3 y^{^{'}} = 2 y