面积 y= 1/2 x,y=1,y= 1/25 x^2

解答

面积

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

\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 = 1 : x = 2

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

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

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

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

\int_{- 5}^{0} 1 - \frac{1}{25} x^{2} 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}{2} x - \frac{1}{25} x^{2} d x = \frac{135}{16}

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

化简

= \frac{721}{48}

n = 5 \sum \infty sin (\frac{1}{n ^{2}})

\frac{d}{d x} (\frac{1}{2} cos (x))

\frac{d}{d x} (x ln (4 y) + x + y)

\int 6 x^{\frac{3}{2}} d x

x \to \infty lim (4^{x} + 5)