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

解

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

\frac{16}{3}

十進法表記

5.33333 \dots

解答ステップ

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

\int_{0}^{2} x^{2} d x = \frac{8}{3}

\int_{0}^{2} 4 d x = 8

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

簡素化

- \frac{8}{3} + 8 : \frac{16}{3}

= \frac{16}{3}