integral from sqrt(8) to-1 of x^2-8

解

\int_{8}^{- 1} x^{2} - 8 d x

- \frac{16 2 + 1}{3} + 162 + 8

十進法表記

22.75161 \dots

解答ステップ

\int_{8}^{- 1} x^{2} - 8 d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{8} x^{2} - 8 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}^{8} x^{2} d x - \int_{- 1}^{8} 8 d x)

\int_{- 1}^{8} x^{2} d x = \frac{16 2 + 1}{3}

\int_{- 1}^{8} 8 d x = 162 + 8

= - (\frac{16 2 + 1}{3} - (162 + 8))

簡素化

= - \frac{16 2 + 1}{3} + 162 + 8