integral from 0 to 1 of x^2(x^3+1)^3

Lösung

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

\frac{5}{4}

Dezimale

1.25

Schritte zur Lösung

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

Multipliziere aus

x^{2} (x^{3} + 1)^{3} : x^{11} + 3 x^{8} + 3 x^{5} + x^{2}

= \int_{0}^{1} x^{11} + 3 x^{8} + 3 x^{5} + x^{2} d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} x^{11} d x + \int_{0}^{1} 3 x^{8} d x + \int_{0}^{1} 3 x^{5} d x + \int_{0}^{1} x^{2} d x

\int_{0}^{1} x^{11} d x = \frac{1}{12}

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

\int_{0}^{1} 3 x^{5} d x = \frac{1}{2}

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

= \frac{1}{12} + \frac{1}{3} + \frac{1}{2} + \frac{1}{3}

Vereinfache

\frac{1}{12} + \frac{1}{3} + \frac{1}{2} + \frac{1}{3} : \frac{5}{4}

= \frac{5}{4}

\frac{d}{d x} ((e^{x})^{x})

in v erse l a pl a ce 1000 \cdot \frac{1}{s ^{2} + 1000 s}

\int \frac{x ^{4}}{4} - \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x + C d x

\int 2 x^{3} d x

d er i v a t i v e \frac{3}{ln ( x )}