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

解

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

\frac{99}{4}

十進法表記

24.75

解答ステップ

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

拡張

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

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

= \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}

簡素化

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

= \frac{99}{4}