Question

高等数学定积分

Answer 1

38(1) y^2 = 2x, x = y^2/2, dx/dy = y,
s = ∫<-1, 1>√[1+(dx/dy)^2] dy = ∫<-1, 1>√(1+y^2)dy
= [y√(1+y^2)]<-1, 1> - ∫<-1, 1>[y^2/√(1+y^2)]dy
= 2√2 - ∫<-1, 1>√(1+y^2)dy + ∫<-1, 1>[1/√(1+y^2)]dy,
2s = 2√2 + ∫<-1, 1>[1/√(1+y^2)]dy (y = tanu)
= 2√2 + ∫<-π/4, π/4>secudu
= 2√2 + [ln(secu+tanu)]<-π/4, π/4>
= 2√2 + 2ln(1+√2), 则 s = √2 + ln(1+√2)。
（2） y = lnx, y' = 1/x,
s = ∫<√3, √8>√[1+(1/x)^2] dx =∫<√3, √8>√(1+x^2)dx/x (x = tanu)
= ∫<π/3, arctan√8>du/[sinu(cosu)^2]
= -∫<π/3, arctan√8>dcosu/[(sinu)^2(cosu)^2]
= - ∫<π/3, arctan√8>dcosu/{[1-(cosu)^2](cosu)^2}
= - ∫<π/3, arctan√8>{1/(cosu)^2 + (1/2)[1/(1-cosu)+1/(1+cosu)]}dcosu
= - [-1/cosu + (1/2)ln{(1+cosu)/(1-cosu)}]<π/3, arctan√8>
= 1+(1/2)(ln3-ln2).
(3) y = ∫<-π/2, x>√costdt , y' = √cosx,
s = ∫<-π/2, π/2> √(1+cosx)dx = ∫<-π/2, π/2> √2cos(x/2)dx
= 2√2[sin(x/2)]<-π/2, π/2> = 4.

Answer 2