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

解答

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

- \frac{9}{4}

十进制

- 2.25

求解步骤

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

乘开

x (2 - x^{2} - x) : 2 x - x^{3} - x^{2}

= \int_{- 2}^{1} 2 x - x^{3} - x^{2} 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}^{1} 2 x d x - \int_{- 2}^{1} x^{3} d x - \int_{- 2}^{1} x^{2} d x

\int_{- 2}^{1} 2 x d x = - 3

\int_{- 2}^{1} x^{3} d x = - \frac{15}{4}

\int_{- 2}^{1} x^{2} d x = 3

= - 3 - (- \frac{15}{4}) - 3

化简

- 3 - (- \frac{15}{4}) - 3 : - \frac{9}{4}

= - \frac{9}{4}

\int_{2}^{6} (6 - y^{2}) y d y

\int_{- 1}^{1} x 5 - x^{2} d x

\int_{0}^{\frac{π}{4}} sec^{3} (θ) d θ

\int_{0}^{4} \frac{1}{x - 1} d x

\int_{2}^{3} 25 d x