Find the exact length of the curve: y = 4 + 4x^3/2, 0 ≤ x ≤ 1.

Solution:

The length of the curve y = f(x) a ≤ x ≤ b is given by:

L = \(\int_{a}^{b}\sqrt{1 + (\frac{\mathrm{d} y}{\mathrm{d} x})^{2}}dx\)

y = 4 + 4x^3/2

dy/dx = 4(3/2)x^1/2

= 6x^1/2

√1 + (dy/dx)² = √1 + (6x^1/2)² = √1 + 36x

Length of the curve L = \(\int_{0}^{1}\sqrt{1 + 36x}dx\)

= \(\frac{2}{3}.\frac{1}{36}[(1+36x)^{3/2}]_{0}^{1}\)

= \(\frac{1}{54}[(1+36(1))^{3/2} - (1+36(0))^{3/2})]\)

= \(\frac{1}{54}[(37)^{3/2} - (1)^{3/2})]\)

= \(\frac{1}{54}[(37)^{3/2} - (1)]\)

Find the exact length of the curve: y = 4 + 4x^3/2, 0 ≤ x ≤ 1.

Summary:

The length of the curve y = 4 + 4x^3/2, 0 ≤ x ≤ 1 is L = (1/54)[(37)^3/2 - 1]