该题是要求不定积分,不定积分是高中数学内容.
F(x)=∫F'(x)dx=∫√(4-x^2)dx=2∫√[1-(x/2)^2]dx
|x/2|≤1,令sint=x/2,则x=2sint
F(x)=2∫√(1-sint^2)d(2sint)=4∫cost^2dt
=4∫costd(sint)=4costsint-4∫sintd(cost)=4costsint+4∫sint^2dt
=4costsint+4∫(1-cost^2)dt=4costsint+4t-4∫cost^2dt
=4costsint+4t-F(x)
F(x)=2costsint+2t+C=sin2t+2t+C
=sin[2arcsin(x/2)]+2arcsin(x/2)+C