作业帮高数:计算下列积分
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/07/08 02:24:42
高数:计算下列积分
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
= ∫(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