integral from 1 to 5 of (x^2+x)

解

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

\frac{160}{3}

十進法表記

53.33333 \dots

解答ステップ

\int_{1}^{5} x^{2} + x 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}^{5} x^{2} d x + \int_{1}^{5} x d x

\int_{1}^{5} x^{2} d x = \frac{124}{3}

\int_{1}^{5} x d x = 12

= \frac{124}{3} + 12

簡素化

\frac{124}{3} + 12 : \frac{160}{3}

= \frac{160}{3}