求定积分上限1下限02(1+1/2x)^1/2dx
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/25 15:39:21
![求定积分上限1下限02(1+1/2x)^1/2dx](/uploads/image/z/11563215-15-5.jpg?t=%E6%B1%82%E5%AE%9A%E7%A7%AF%E5%88%86%E4%B8%8A%E9%99%901%E4%B8%8B%E9%99%9002%281%2B1%2F2x%29%5E1%2F2dx)
求定积分上限1下限02(1+1/2x)^1/2dx
求定积分上限1下限02(1+1/2x)^1/2dx
求定积分上限1下限02(1+1/2x)^1/2dx
原式=∫(+∞→1/2)2√(1+u)*(-1)/(2u^2)*du(设u=1/(2x))
=√6+(1/2)∫(√(3/2)→+∞)1/[(t^2-1)*t]*2tdt(设√(1+u)=t)=√6+(1/2)ln(5+√6)
令u² = 1 + 1/2x
x = 1/[2(u² - 1)],dx = - u/(u² - 1)² du
∫(0,1) 2√[1 + 1/(2x)] dx
= 2∫(+∞,√(3/2)) - u²/(u² - 1)² du
令u = secθ,du = secθtanθ dθ
= 2∫...
全部展开
令u² = 1 + 1/2x
x = 1/[2(u² - 1)],dx = - u/(u² - 1)² du
∫(0,1) 2√[1 + 1/(2x)] dx
= 2∫(+∞,√(3/2)) - u²/(u² - 1)² du
令u = secθ,du = secθtanθ dθ
= 2∫(arccos√(2/3),π/2) (sec²θ*secθtanθ dθ)/(sec²θ - 1)²
= 2∫(arccos√(2/3),π/2) (sec³θtanθ dθ)/tan⁴θ
= 2∫(arccos√(2/3),π/2) csc³θ dθ
= 2[(1/2)ln|cscθ - cotθ| - (1/2)cscθcotθ] |(arccos√(2/3),π/2]
= [ln(1 + 0) - (1/2)(1)(0)] - [ln(√3 - √2) - (√3)(√2)]
= √6 - ln(√3 - √2) ≈ 3.59571
收起