面积 y=sqrt(x),y=x-2,y=0

解答

面积

y = x, y = x - 2, y = 0

\frac{10}{3}

十进制

3.33333 \dots

求解步骤

要找到相交点，解

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

x = x - 2 : x = 4

x = 0 : x = 0

x - 2 = 0 : x = 2

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

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

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

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

= \frac{4 2}{3} + \frac{16 - 4 2}{3} - 2

化简

= \frac{10}{3}

t an g e n t f (x) = 3 x^{- 2}, at x = 4

\int_{0}^{π^{2}} sin (x) d x

d er i v a t i v e 3 x^{\frac{1}{3}} - 5 x^{\frac{4}{3}}

\int \frac{sin ^{7} ( 5 x )}{cos ^{4} ( 5 x )} d x

\int t arcsec (t) d t