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

Lösung

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

\frac{40}{3} f

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Multipliziere aus

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

= f \cdot \int_{0}^{2} x^{2} + x^{5} 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

= f (\int_{0}^{2} x^{2} d x + \int_{0}^{2} x^{5} d x)

\int_{0}^{2} x^{2} d x = \frac{8}{3}

\int_{0}^{2} x^{5} d x = \frac{32}{3}

= f (\frac{8}{3} + \frac{32}{3})

Füge

\frac{8}{3} + \frac{32}{3}

zusammen:

\frac{40}{3}

= \frac{40}{3} f

\int_{- π}^{π} 1 d x

\int_{1}^{l n (8)} 6 x + e^{x} d x

\int_{0}^{\frac{π}{2}} \frac{1}{2 + sin ( x )} d x

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

\int_{- 2}^{4} (1 - x - 4 x^{3} + 2 x^{5}) d x