integral from 1 to 4 of x(1+x)

解

\int_{1}^{4} x (1 + x) d x

\frac{57}{2}

十進法表記

28.5

解答ステップ

\int_{1}^{4} x (1 + x) d x

拡張

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

= \int_{1}^{4} x + x^{2} 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_{1}^{4} x d x + \int_{1}^{4} x^{2} d x

\int_{1}^{4} x d x = \frac{15}{2}

\int_{1}^{4} x^{2} d x = 21

= \frac{15}{2} + 21

簡素化

\frac{15}{2} + 21 : \frac{57}{2}

= \frac{57}{2}