面积 y=10x,y=x^2-11,x

解答

面积

y = 10 x, y = x^{2} - 11, x

\frac{531}{2}

十进制

265.5

求解步骤

要找到相交点，解

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

10 x = x^{2} - 11 : x = 11, x = - 1

10 x = x : x = 0

x^{2} - 11 = x : x = \frac{1 + 3 5}{2}, x = \frac{1 - 3 5}{2}

= \int_{\frac{1 - 3 5}{2}}^{- 1} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{- 1}^{0} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{0}^{\frac{1 + 3 5}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{\frac{1 + 3 5}{2}}^{11} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{\frac{1 - 3 5}{2}}^{- 1} x^{2} - 11 - x d x + \int_{- 1}^{0} ∣ 10 x - x ∣ d x + \int_{0}^{\frac{1 + 3 5}{2}} ∣ 10 x - x ∣ d x + \int_{\frac{1 + 3 5}{2}}^{11} x^{2} - 11 - 10 x d x

\int_{\frac{1 - 3 5}{2}}^{- 1} x^{2} - 11 - x d x = \frac{45 5 - 63}{4}

\int_{- 1}^{0} ∣ 10 x - x ∣ d x = \frac{9}{2}

\int_{0}^{\frac{1 + 3 5}{2}} ∣ 10 x - x ∣ d x = \frac{9 ( 23 + 3 5 )}{4}

\int_{\frac{1 + 3 5}{2}}^{11} x^{2} - 11 - 10 x d x = 225 - 185

= \frac{45 5 - 63}{4} + \frac{9}{2} + \frac{9 ( 23 + 3 5 )}{4} + 225 - 185

化简

= \frac{531}{2}

x^{^{''}} + 2 x^{^{'}} + 2 x = 2

\frac{d}{d x} (- 2 ln (cos (x)))

\int x \cdot e^{2 x^{2}} d x

\int_{1}^{2} - x^{2} + 3 x - 2 d x

2 x^{2} \frac{d y}{d x} + 2 x y = y^{3}