面积 (x+2)^3,2x^2-8x+8,0

解答

面积

(x + 2)^{3}, 2 x^{2} - 8 x + 8, 0

\frac{28}{3}

十进制

9.33333 \dots

求解步骤

要找到相交点，解

f_{i} (x) = f_{j} (x)

(x + 2)^{3} = 2 x^{2} - 8 x + 8 : x = 0

(x + 2)^{3} = 0 : x = - 2

2 x^{2} - 8 x + 8 = 0 : x = 2

= \int_{- 2}^{0} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{0}^{2} ∣ f_{3} (x) - f_{2} (x) ∣ d x

= \int_{- 2}^{0} (x + 2)^{3} - 0 d x + \int_{0}^{2} 0 - (2 x^{2} - 8 x + 8) d x

\int_{- 2}^{0} (x + 2)^{3} - 0 d x = 4

\int_{0}^{2} 0 - (2 x^{2} - 8 x + 8) d x = \frac{16}{3}

= 4 + \frac{16}{3}

化简

= \frac{28}{3}

x \to 1 lim (x + 4 - 12)

t an g e n t f (x) = x + 2, at x = 1

d er i v a t i v e f (x) = lo g_{4} (3 x^{2} + 1)

x \to 14 - lim (15 x + ln (14 - x))

l a pl a ce t r an s f or m (1 + e^{- t})^{2}