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

解

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

\frac{40}{3} f

解答ステップ

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

定数を除く:

\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

拡張

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

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

総和規則を適用する:

\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})

結合

\frac{8}{3} + \frac{32}{3} : \frac{40}{3}

= \frac{40}{3} f