∫[from −2 to −4]√(x²−4) dx
Hence, show that
−
4√3+2ln(2−√3)
≤
∫[from −2 to −4]
√(x²−4)
dx ≤ −
4√3+2ln(2−√3)
8
x³
64
Evaluate
∫[from −2 to −4]
√(x²−4)
dx
x³
Check whether the answer satisfy the above inequality.
更新1:
Sorry the limit should be from -4 to -2
更新2:
One interesting thing to ask is that why we are not using principal values of arcsecant, i.e. x=-2 =>pi x=-4 =>2pi/3 but we should use x=-2 =>pi x=-4 =>2pi/3 instead in doing transformation of boundaries?
更新3:
Correction: One interesting thing to ask is that why we are not using principal values of arcsecant, i.e. x=-2 =>pi x=-4 =>2pi/3 but we should use x=-2 =>pi x=-4 =>4pi/3 instead in doing transformation of boundaries?
