面积 y=(x-2)^2,y=4-x,y=2sqrt(x-2)

解答

面积

y = (x - 2)^{2}, y = 4 - x, y = 2 x - 2

- \frac{15}{2} + \frac{- 8 ( 4 - 2 3 ) ^{\frac{3}{2}} + 14}{3} + 83 - 5.66666 \dots

十进制

4.31026 \dots

求解步骤

要找到相交点，解

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

(x - 2)^{2} = 4 - x : x = 3, x = 0

(x - 2)^{2} = 2 x - 2 : x = 2, x \approx 3.58740 \dots

4 - x = 2 x - 2 : x = 2 (3 - 3)

= \int_{0}^{2} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{2}^{2 (3 - 3)} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{2 (3 - 3)}^{3} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{3}^{3.58740 \dots} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{0}^{2} (x - 2)^{2} - (4 - x) d x + \int_{2}^{2 (3 - 3)} 4 - x - 2 x - 2 d x + \int_{2 (3 - 3)}^{3} 4 - x - 2 x - 2 d x + \int_{3}^{3.58740 \dots} (x - 2)^{2} - 2 x - 2 d x

\int_{0}^{2} (x - 2)^{2} - (4 - x) d x = \frac{10}{3}

\int_{2}^{2 (3 - 3)} 4 - x - 2 x - 2 d x = - \frac{4 ( 4 - 2 3 ) ^{\frac{3}{2}}}{3} + 43 - 6

\int_{2 (3 - 3)}^{3} 4 - x - 2 x - 2 d x = 43 - \frac{15}{2} + \frac{2 ( - 2 ( 4 - 2 3 ) ^{\frac{3}{2}} + 2 )}{3}

\int_{3}^{3.58740 \dots} (x - 2)^{2} - 2 x - 2 d x = 0.33333 \dots

= \frac{10}{3} - \frac{4 ( 4 - 2 3 ) ^{\frac{3}{2}}}{3} + 43 - 6 + 43 - \frac{15}{2} + \frac{2 ( - 2 ( 4 - 2 3 ) ^{\frac{3}{2}} + 2 )}{3} + 0.33333 \dots

化简

= - \frac{15}{2} + \frac{- 8 ( 4 - 2 3 ) ^{\frac{3}{2}} + 14}{3} + 83 - 5.66666 \dots

\frac{d}{d x} (\frac{75}{12 + x ^{2} + y ^{2}})

x \to 3 lim (x 18 - x^{2})

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

d er i v a t i v e y = 2 e^{x} cos (x)

in v erse l a pl a ce \frac{( \frac{1}{2} )}{s}