面积 y=sqrt(2x),y=3x-4,y=0

解答

面积

y = 2 x, y = 3 x - 4, y = 0

2

求解步骤

要找到相交点，解

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

2 x = 3 x - 4 : x = 2

2 x = 0 : x = 0

3 x - 4 = 0 : x = \frac{4}{3}

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

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

\int_{0}^{\frac{4}{3}} 2 x - 0 d x = \frac{16 2}{9 3}

\int_{\frac{4}{3}}^{2} 3 x - 4 - 2 x d x = \frac{24 3 - 16 2}{9 3} - \frac{2}{3}

= \frac{16 2}{9 3} + \frac{24 3 - 16 2}{9 3} - \frac{2}{3}

化简

= 2

x \to 2 lim (\frac{2 x}{3 x + 1})

\frac{\partial}{\partial y} (y^{3} cos (2 x))

k = 1 \sum \infty \frac{( - 1 ) ^{k} \cdot ( 1 ) ^{(3 \cdot k)}}{k}

\int x y (x + y) d y

n = 0 \sum \infty \frac{2 ^{n}}{n !}