定积分(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0)^1sqrt (2x-{x)^2}dx=( )(int )_(0)^1sqrt (2x-{x)^2}dx=( )

(8)(int )_(1)^2sqrt (2x-{x)^2}dx=__________________.(8)__________________.

[例5] 积分 (int )_(0)^2dx(int )_(0)^sqrt (2x-{x^2)}sqrt ({x)^2+(y)^2}dy= __

269 int_(0)^1dyint_(y)^1sqrt(x^2)-y^(2)dx的值为A. $\frac{\pi}{3}$.B. $\frac{\pi}{6}

(int )_(0)^2xsqrt (2x-{x)^2}dx= __

求不定积分int dfrac (1)(2x)sqrt (ln x)dx=（）.int dfrac (1)(2x)sqrt (ln x)dx=int dfrac

(int )_(0)^1dfrac (x+2)(sqrt {x+1)}dx=( )(int )_(0)^1dfrac (x+2)(sqrt {x+1)}dx=

8.设f(x)满足等式 (x)-f(x)=sqrt (2x-{x)^2}, 且 (1)=4, 则 (int )_(0)^1f(x)dx= __

(1) (int )_(0)^a(x)^2sqrt ({a)^2-(x)^2}dx(agt 0);

(8) (int )_(0)^1(x)^2sqrt (1-{x)^2}dx :

(int )_(0)^sqrt (2)sqrt (2-{x)^2}dx