I = ∫
= ∫<0, 2>(1+9x^4)dx = [x+(9/5)x^5]<0, 2> = 2+(9/5)32 = 298/5
高数题目求解，如图
y = x^3, y' = 3x^2, ds = √(1+y'^2)dx = √(1+9x^4)dx
I = ∫f(x,y)ds = ∫<0, 2>√[1+9x(x^3)]√(1+9x^4)dx
= ∫<0, 2>(1+9x^4)dx = [x+(9/5)x^5]<0, 2> = 2+(9/5)32 = 298/5
I = ∫
= ∫<0, 2>(1+9x^4)dx = [x+(9/5)x^5]<0, 2> = 2+(9/5)32 = 298/5
