integral from 2 to 0 of-4x^4+8x^3

解

\int_{2}^{0} - 4 x^{4} + 8 x^{3} d x

- \frac{32}{5}

十進法表記

- 6.4

解答ステップ

\int_{2}^{0} - 4 x^{4} + 8 x^{3} d x

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

= - \int_{0}^{2} - 4 x^{4} + 8 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}^{2} 4 x^{4} d x + \int_{0}^{2} 8 x^{3} d x)

\int_{0}^{2} 4 x^{4} d x = \frac{128}{5}

\int_{0}^{2} 8 x^{3} d x = 32

= - (- \frac{128}{5} + 32)

結合

- \frac{128}{5} + 32 : \frac{32}{5}

= - \frac{32}{5}