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

Lösung

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

\frac{99}{4}

Dezimale

24.75

Schritte zur Lösung

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

Multipliziere aus

x (1 + x^{2}) : x + x^{3}

= \int_{0}^{3} x + x^{3} 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}^{3} x d x + \int_{0}^{3} x^{3} d x

\int_{0}^{3} x d x = \frac{9}{2}

\int_{0}^{3} x^{3} d x = \frac{81}{4}

= \frac{9}{2} + \frac{81}{4}

Vereinfache

\frac{9}{2} + \frac{81}{4} : \frac{99}{4}

= \frac{99}{4}

\int_{9}^{36} \frac{1}{x} d x

\int_{3}^{3} \frac{8}{9 + x ^{2}} d x

\int_{0}^{1} \frac{1}{2} x + \frac{1}{2} - x^{2} d x

\int_{0}^{5} 2.5 x d x

\int_{x}^{x^{3}} \frac{y}{x} d y