integral from 0 to 3 of (x-1)(x-2)(x-3)

解答

\int_{0}^{3} (x - 1) (x - 2) (x - 3) d x

- \frac{9}{4}

十进制

- 2.25

求解步骤

\int_{0}^{3} (x - 1) (x - 2) (x - 3) d x

乘开

(x - 1) (x - 2) (x - 3) : x^{3} - 6 x^{2} + 11 x - 6

= \int_{0}^{3} x^{3} - 6 x^{2} + 11 x - 6 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_{0}^{3} x^{3} d x - \int_{0}^{3} 6 x^{2} d x + \int_{0}^{3} 11 x d x - \int_{0}^{3} 6 d x

\int_{0}^{3} x^{3} d x = \frac{81}{4}

\int_{0}^{3} 6 x^{2} d x = 54

\int_{0}^{3} 11 x d x = \frac{99}{2}

\int_{0}^{3} 6 d x = 18

= \frac{81}{4} - 54 + \frac{99}{2} - 18

化简

\frac{81}{4} - 54 + \frac{99}{2} - 18 : - \frac{9}{4}

= - \frac{9}{4}

s l o p e in t erce pt (- 7, - 9), (0, - 7)

\int (4 - y^{2})^{\frac{1}{2}} d y

\frac{d}{d x} (sin (1 - x) cos^{3} (x))

d er i v a t i v e f (x) = \frac{e ^{x}}{2 + 5 x}

x + \frac{y d y}{d x} = 0