integral from-2 to 2 of (-5x^2-4x)^3

解答

\int_{- 2}^{2} (- 5 x^{2} - 4 x)^{3} d x

- \frac{53504}{7}

十进制

- 7643.42857 \dots

求解步骤

\int_{- 2}^{2} (- 5 x^{2} - 4 x)^{3} d x

乘开

(- 5 x^{2} - 4 x)^{3} : - 125 x^{6} - 300 x^{5} - 240 x^{4} - 64 x^{3}

= \int_{- 2}^{2} - 125 x^{6} - 300 x^{5} - 240 x^{4} - 64 x^{3} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- 2}^{2} 125 x^{6} d x - \int_{- 2}^{2} 300 x^{5} d x - \int_{- 2}^{2} 240 x^{4} d x - \int_{- 2}^{2} 64 x^{3} d x

\int_{- 2}^{2} 125 x^{6} d x = \frac{32000}{7}

\int_{- 2}^{2} 300 x^{5} d x = 0

\int_{- 2}^{2} 240 x^{4} d x = 3072

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

= - \frac{32000}{7} - 0 - 3072 - 0

化简

- \frac{32000}{7} - 0 - 3072 - 0 : - \frac{53504}{7}

= - \frac{53504}{7}

\int_{0}^{\infty} 9 e^{- 3 x} d x

\int_{1}^{2 e} (\frac{2}{x} + e^{x}) d x

\int_{1}^{x} (t^{2}) d t

\int_{0}^{4} x (4 - x) d x

\int_{0}^{20} ∣ x - 10 ∣ d x