求定积分

Question

定积分f xcox dx X的取值为-1到1
定积分f 根号（4-x平方） dx X的取值为0到1

Answer 1

∫xcosxdx
=∫xdsinx
=xsinx-∫sinxdx
=xsinx+cosx

∫<1，-1>xcosxdx
=xsinx+cosx|<1，-1>
=sin1+cos1-[-sin(-1)+cos(-1)]
=0

令x=2sint，dx=2costdt
x上限1、下限0，则sint上限1/2、下限0，则t上限π/6、下限0，则2t上限π/3、下限0

∫√(4-x²)dx
=2∫cost√(4-4sin²t)dt
=4∫cost√cos²tdt
=4∫cos²tdt
=2∫1+cos(2t)dt
=∫1+cos(2t)d(2t)
=2t+sin(2t)

∫√(4-x²)dx
=2t+sin(2t)|<π/3，0>
=π/3+√3/2本回答被提问者采纳

Answer 2

Answer 3