面积 sqrt(x),x-2,y=0

解答

面积

x, x - 2, y = 0

\frac{10}{3}

十进制

3.33333 \dots

求解步骤

要找到相交点，解

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

0 = x : x = 0

0 = x - 2 : x = 2

x = x - 2 : x = 4

= \int_{0}^{2} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{2}^{4} ∣ f_{2} (x) - f_{3} (x) ∣ d x

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

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

\int_{2}^{4} x - (x - 2) d x = \frac{16 - 4 2}{3} - 2

= \frac{4 2}{3} + \frac{16 - 4 2}{3} - 2

化简

= \frac{10}{3}

x \to 0 lim (\frac{sin ( k x )}{x})

\int \frac{3 r}{1 - r ^{2}} d r

\int cos^{2} (4 x) d x

d er i v a t i v e y = (\frac{x}{3.3} + \frac{3.3}{x}) (x^{2} + 1)

x \to + 2 lim (2 + 3)