面积 1/x ,x, 1/4 x

解答

面积

\frac{1}{x}, x, \frac{1}{4} x

2 ln (2)

十进制

1.38629 \dots

求解步骤

要找到相交点，解

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

\frac{1}{x} = x : x = 1, x = - 1

\frac{1}{x} = \frac{1}{4} x : x = 2, x = - 2

x = \frac{1}{4} x : x = 0

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

= \int_{- 2}^{- 1} \frac{1}{x} - \frac{1}{4} x d x + \int_{- 1}^{0} \frac{1}{4} x - x d x + \int_{0}^{1} x - \frac{1}{4} x d x + \int_{1}^{2} \frac{1}{x} - \frac{1}{4} x d x

\int_{- 2}^{- 1} \frac{1}{x} - \frac{1}{4} x d x = ln (2) - \frac{3}{8}

\int_{- 1}^{0} \frac{1}{4} x - x d x = \frac{3}{8}

\int_{0}^{1} x - \frac{1}{4} x d x = \frac{3}{8}

\int_{1}^{2} \frac{1}{x} - \frac{1}{4} x d x = ln (2) - \frac{3}{8}

= ln (2) - \frac{3}{8} + \frac{3}{8} + \frac{3}{8} + ln (2) - \frac{3}{8}

化简

= 2 ln (2)

\int \frac{5 x}{x ^{2} + 4 x + 4} d x

t a y l or f (x) = \frac{1}{3 - 2 x}

s l o p e (3, 5), (0, - 1)

\int 2 sin^{2} (\frac{θ}{2}) d θ

t a y l or \frac{x}{( 1 - x ^{2} )}