面积-4x+x^2+10,9-2x,2x+1

解答

面积

- 4 x + x^{2} + 10, 9 - 2 x, 2 x + 1

\frac{2}{3}

十进制

0.66666 \dots

求解步骤

要找到相交点，解

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

- 4 x + x^{2} + 10 = 9 - 2 x : x = 1

- 4 x + x^{2} + 10 = 2 x + 1 : x = 3

9 - 2 x = 2 x + 1 : x = 2

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

= \int_{1}^{2} - 4 x + x^{2} + 10 - (9 - 2 x) d x + \int_{2}^{3} - 4 x + x^{2} + 10 - (2 x + 1) d x

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

\int_{2}^{3} - 4 x + x^{2} + 10 - (2 x + 1) d x = \frac{1}{3}

= \frac{1}{3} + \frac{1}{3}

化简

= \frac{2}{3}

d er i v a t i v e y = x - 2 3 x

d er i v a t i v e (2) (e^{3 x})

im pl i c i t \frac{d y}{d x}, x^{2} + x y + y^{2} = 7

\int_{0}^{\frac{π}{3}} 2 sec^{2} (x) d x

a re a f (x) = x + 3, g (x) = \frac{1}{2} x + 3