Question

定积分题目。

积分0到1 根号下4减x平方 dx=

Answer 1

解：设x=2sinz，则dx=2coszdz
于是，∫<0,1>√(4-x²)dx=∫<0,π/6>2cosz*2coszdz
=4∫<0,π/6>cos²zdz
=2∫<0,π/6>[1+cos(2z)]dz
=2[z+sin(2z)/2]│<0,π/6>
=2[π/6+sin(π/3)/2]
=2(π/6+√3/4)
=π/3+√3/2。

Answer 2

∫<0,1>√(4-x²)dx
设x=2sinθ
=∫<0,π/6>2cosθd(2sinθ)
=∫<0,π/6>4cos²θdθ
=∫<0,π/6>2(cos2θ+1)dθ
=∫<0,π/6>cos2θd(2θ)+2∫<0,π/6>dθ
=sin(2θ)|<0,π/6>+π/3
=√3/2+π/3