计算定积分 (∫)_(0)^2sqrt(4-(x)^2)dx= ____ .计算定积分${∫}_{0}^{2}$$\sqrt{4-{x}^{2}}$dx= __
[题目]-|||-定积分 (int )_(-2)^2sqrt (4-{x)^2}dx= __
(int )_(-2)^2(sqrt (4-{x)^2}+x)dx 等于( )等于( ) A. 0
(int )_(0)^a(x)^2sqrt ({a)^2-(x)^2}dx(agt 0)
(1) (int )_(0)^a(x)^2sqrt ({a)^2-(x)^2}dx(agt 0);
(3)(int )_(0)^a(x)^2sqrt ({a)^2-(x)^2}dx(agt 0);
(({int )_(-2)^2}(x-2)sqrt (4-{x)^2}dx
(int )_(-2)^2(x-2)sqrt (4-{x)^2}dx
2.计算下列定积分.-|||-(3) (int )_(-2)^2((x)^2sqrt (4-{x)^2}+x(cos )^5x)dx.
(8) (int )_(0)^1(x)^2sqrt (1-{x)^2}dx :