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

解

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

\frac{20}{3}

十進法表記

6.66666 \dots

解答ステップ

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

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

\int_{0}^{4} 2 x d x = 16

\int_{0}^{4} 3 d x = 12

= - \frac{64}{3} + 16 + 12

簡素化

- \frac{64}{3} + 16 + 12 : \frac{20}{3}

= \frac{20}{3}