面积 20x-50,y=0,2x^2

解答

面积

20 x - 50, y = 0, 2 x^{2}

\frac{125}{6}

十进制

20.83333 \dots

求解步骤

要找到相交点，解

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

0 = 20 x - 50 : x = \frac{5}{2}

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

20 x - 50 = 2 x^{2} : x = 5

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

= \int_{0}^{\frac{5}{2}} 0 - 2 x^{2} d x + \int_{\frac{5}{2}}^{5} 2 x^{2} - (20 x - 50) d x

\int_{0}^{\frac{5}{2}} 0 - 2 x^{2} d x = \frac{125}{12}

\int_{\frac{5}{2}}^{5} 2 x^{2} - (20 x - 50) d x = \frac{125}{12}

= \frac{125}{12} + \frac{125}{12}

化简

= \frac{125}{6}

\int (4 - r^{2}) r d r

d er i v a t i v e f (x) = 9 x

n \to \infty lim (n^{n})

\frac{d}{d x} (8 x + 2)

\int 2^{2 x} d x