The length L of the curve y = f(x) from x = a to x = b is
L = ∫_a^b √(1 + {f'(x)}²) dx
For the parametric representation x = x(t), y = y(t),
L = ∫_α^β √((dx/dt)² + (dy/dt)²) dt
Example ① The length of y = x^{3/2} from x=0 to x=4.
Solution: y' = (3/2)x^{1/2}、1 + (y')² = 1 + (9/4)x。
L = ∫_0^4 √(1+9x/4) dx。u=1+9x/4、du=(9/4)dx → dx=4du/9。区間 1→10。
L = (4/9)·[2u^{3/2}/3]_1^{10} = (8/27)(10√10 − 1)
Solution: y' = (3/2)x^{1/2}、1 + (y')² = 1 + (9/4)x。
L = ∫_0^4 √(1+9x/4) dx。u=1+9x/4、du=(9/4)dx → dx=4du/9。区間 1→10。
L = (4/9)·[2u^{3/2}/3]_1^{10} = (8/27)(10√10 − 1)
Key Point
- For integrating √(1+(y')²), substitution is central
- For parametric equations, take the square root of the sum of the squares of both derivatives
Practice Problems
- The length of the circle x=r cos t、y=r sin t（0≦t≦2π）.
Solution and Explanation
- Answer: 2πr
√((−r sin t)²+(r cos t)²) = r。L=∫_0^{2π} r dt=2πr。