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

解答

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

\frac{15}{2}

十进制

7.5

求解步骤

\int_{- 1}^{2} 3 x + 1 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_{- 1}^{2} 3 x d x + \int_{- 1}^{2} 1 d x

\int_{- 1}^{2} 3 x d x = \frac{9}{2}

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

= \frac{9}{2} + 3

化简

\frac{9}{2} + 3 : \frac{15}{2}

= \frac{15}{2}

in v erse l a pl a ce \frac{x ^{2} + 1}{x ^{2} - x - 6}

d er i v a t i v e f (x) = 8 sin^{6} (x)

x \to 4 + lim (\frac{1}{∣ 4 - x ∣})

s l o p e y = x^{3}, at x = 5

\frac{d}{d x} (3 x^{4} - 2 x^{3} - 12 x^{2} + 18 x + 15)