面积 y= 1/2 x,y= 1/25 x^2,y=1

解答

面积

y = \frac{1}{2} x, y = \frac{1}{25} x^{2}, y = 1

\frac{721}{48}

十进制

15.02083 \dots

求解步骤

要找到相交点，解

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

\frac{1}{2} x = \frac{1}{25} x^{2} : x = \frac{25}{2}, x = 0

\frac{1}{2} x = 1 : x = 2

\frac{1}{25} x^{2} = 1 : x = 5, x = - 5

= \int_{- 5}^{0} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{0}^{2} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{2}^{5} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{5}^{\frac{25}{2}} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{- 5}^{0} \frac{1}{25} x^{2} - 1 d x + \int_{0}^{2} \frac{1}{2} x - 1 d x + \int_{2}^{5} \frac{1}{2} x - 1 d x + \int_{5}^{\frac{25}{2}} \frac{1}{25} x^{2} - \frac{1}{2} x d x

\int_{- 5}^{0} \frac{1}{25} x^{2} - 1 d x = \frac{10}{3}

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

\int_{2}^{5} \frac{1}{2} x - 1 d x = \frac{9}{4}

\int_{5}^{\frac{25}{2}} \frac{1}{25} x^{2} - \frac{1}{2} x d x = \frac{135}{16}

= \frac{10}{3} + 1 + \frac{9}{4} + \frac{135}{16}

化简

= \frac{721}{48}

x \to 5 lim (\frac{2 x}{x - 5})

x \to 9 lim (f (x))

\frac{d}{d x} (10 x + 7 x)

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

\frac{d}{d x} (2 x^{4} - 5 x^{3} + x^{2})