integral from 0 to 1 of (u+2)(u-3)

解答

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

- \frac{37}{6}

十进制

- 6.16666 \dots

求解步骤

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

乘开

(u + 2) (u - 3) : u^{2} - u - 6

= \int_{0}^{1} u^{2} - u - 6 d u

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}^{1} u^{2} d u - \int_{0}^{1} u d u - \int_{0}^{1} 6 d u

\int_{0}^{1} u^{2} d u = \frac{1}{3}

\int_{0}^{1} u d u = \frac{1}{2}

\int_{0}^{1} 6 d u = 6

= \frac{1}{3} - \frac{1}{2} - 6

化简

\frac{1}{3} - \frac{1}{2} - 6 : - \frac{37}{6}

= - \frac{37}{6}

\int \frac{x ^{2} ln ^{3} ( 1 + x ^{3} )}{1 + x ^{3}} d x

x \to 0 lim (\frac{1}{2 ^{x}})

p a r i t y sin (tan (2 x))

\int 3 x (x^{2} - \frac{1}{3}) d x

t an g e n t f (x) = \frac{x}{x + 1}, at x = - 2