Question

高数:计算下列积分

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Answer 1

1。L = ∫(0->1) √(2x-x²) dx
= ∫(0->1) √[(x-0)(2-x)] dx
Let x = 2sin²t，dx = 4sintcost，2-x = 2cos²t
x = 0，t = 0，x = 1，t = π/4
L = ∫(0->π/4) √(2²sin²tcos²t) * 4sintcost dt
= 8∫(0->π/4) sin²tcos²t dt
= 8∫(0->π/4) (1/2 * sin2t)² dt
= 2∫(0->π/4) sin²2t dt
= ∫(0->π/4) (1-cos4t) dt
= t - (1/4)sin(4t)
= [π/4 - (1/4)sin(π)] - [0 - 0]
= π/4

2。L = ∫(0->π/2) √(1-sin2x) dx
= 2∫(0->π/4) √(1-sin2x) dx，关于x = π/4对称
u = 1 - sin2x，x = (1/2)arcsin(1-u)，dx = 1/2 * -1/√u * 1/√(2-u) du
x = 0，u = 1，x = π/4，u = 0
L = 2∫(1->0) √u * 1/2 * -1/√u * 1/√(2-u) du
= ∫(0->1) 1/√(2-u) du
= -∫(0->1) 1/√(2-u) d(2-u)
= -2√(2-u)
= -2(1-√2)
= 2(√2-1)

楼上要注意一点，y = √(1-sin2x)是恒大于0的，所以其定积分也必定大于0

Answer 2

∫(0 1)(2x-x^2)^0.5dx
=∫(0 1)(1-1+2x-x^2)^0.5dx
=-∫(0 1)(1-(1-x)^2)^0.5d(1-x) sin(t)=(1-x) x(0 1) t(π/2 0)
=-∫(π/2 0)(1-sin^2(t))^0.5dsin(t)
=-∫(π/2 0)cos(t)^2dt
=-t/2-sin(2t)/4|(π/2 0)
=π/4

∫(0 π/2)(1-sin(2x))^0.5dx
=∫(0 π/2)(sin^2(x)-2sin(x)cos(x)+cos^2(x))^0.5dx
=∫(0 π/2)|(sin(x)-cos(x))|dx
=∫(0 π/4)(-sin(x)+cos(x))dx+∫(π/4 π/2)(sin(x)-cos(x))dx
=(cos(x)+sin(x))|(0 π/4)+(-cos(x)-sin(x))|(π/4 π/2)
=2^0.5/2+2^0.5/2-1+-1+2^0.5/2+2^0.5/2
=2*2^0.5-2