integral from 1 to 5 of x^3-10x^2+29x-20

解

\int_{1}^{5} x^{3} - 10 x^{2} + 29 x - 20 d x

\frac{32}{3}

十進法表記

10.66666 \dots

解答ステップ

\int_{1}^{5} x^{3} - 10 x^{2} + 29 x - 20 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^{3} d x - \int_{1}^{5} 10 x^{2} d x + \int_{1}^{5} 29 x d x - \int_{1}^{5} 20 d x

\int_{1}^{5} x^{3} d x = 156

\int_{1}^{5} 10 x^{2} d x = \frac{1240}{3}

\int_{1}^{5} 29 x d x = 348

\int_{1}^{5} 20 d x = 80

= 156 - \frac{1240}{3} + 348 - 80

簡素化

156 - \frac{1240}{3} + 348 - 80 : \frac{32}{3}

= \frac{32}{3}