integral from 0 to 4 of x^3+x-x^2

解

\int_{0}^{4} x^{3} + x - x^{2} d x

\frac{152}{3}

十進法表記

50.66666 \dots

解答ステップ

\int_{0}^{4} x^{3} + 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_{0}^{4} x^{3} d x + \int_{0}^{4} x d x - \int_{0}^{4} x^{2} d x

\int_{0}^{4} x^{3} d x = 64

\int_{0}^{4} x d x = 8

\int_{0}^{4} x^{2} d x = \frac{64}{3}

= 64 + 8 - \frac{64}{3}

簡素化

64 + 8 - \frac{64}{3} : \frac{152}{3}

= \frac{152}{3}