Question

根号(r^2 - x^2),在区间 ( -r ,r)内的定积分是?

Answer 1

√(r²-x²)是偶函数,所以 ( -r ,r)∫√(r²-x²)dx =(0,r)2∫√(r²-x²)dx
令x=rsint t∈[-π/2,π/2] 那么区间变为在(0,π/2)内的定积分
=(0,π/2)2∫√(r²-r²sin²t)d(rsint)
=(0,π/2)2∫r²cos²tdt
=(0,π/2)2∫r²(cos2t+1)/2 dt
=(0,π/2)∫r²cos2t/2d(2t)+(0,π/2)∫r²dt
=r²{(0,π/2)[sin2t]+π/2}
=πr²/2
不懂的地方请追问