面积 y=4x^2,y=6x^2,y=6-5x

解答

面积

y = 4 x^{2}, y = 6 x^{2}, y = 6 - 5 x

\frac{3191}{864}

十进制

3.69328 \dots

求解步骤

要找到相交点，解

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

4 x^{2} = 6 x^{2} : x = 0

4 x^{2} = 6 - 5 x : x = \frac{3}{4}, x = - 2

6 x^{2} = 6 - 5 x : x = \frac{2}{3}, x = - \frac{3}{2}

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

= \int_{- 2}^{- \frac{3}{2}} 4 x^{2} - (6 - 5 x) d x + \int_{- \frac{3}{2}}^{0} 6 x^{2} - 4 x^{2} d x + \int_{0}^{\frac{2}{3}} 4 x^{2} - 6 x^{2} d x + \int_{\frac{2}{3}}^{\frac{3}{4}} 4 x^{2} - (6 - 5 x) d x

\int_{- 2}^{- \frac{3}{2}} 4 x^{2} - (6 - 5 x) d x = \frac{29}{24}

\int_{- \frac{3}{2}}^{0} 6 x^{2} - 4 x^{2} d x = \frac{9}{4}

\int_{0}^{\frac{2}{3}} 4 x^{2} - 6 x^{2} d x = \frac{16}{81}

\int_{\frac{2}{3}}^{\frac{3}{4}} 4 x^{2} - (6 - 5 x) d x = \frac{97}{2592}

= \frac{29}{24} + \frac{9}{4} + \frac{16}{81} + \frac{97}{2592}

化简

= \frac{3191}{864}

y = (x^{x} + 1)^{2}

\frac{d}{d θ} (- \frac{n}{2} \cdot ln (2 π) - \frac{n}{2} \cdot ln (θ^{2}))

x \to 0 lim (\frac{7}{2 x})

\int 100 - 2 x d x

\int_{1}^{2} (2 x - 5)^{3} d x