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

Solution:

Given, function is y = 1 + 2x^3/2

Differentiating the function with respect to x,

dy/dx = d(1 + 2x^3/2)/dx

dy/dx = 0+ (3/2)(2)(x^1/2)

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

To find arc length,

Arc length, L = \(\int_{a}^{b}\)√[1 + (dy/dx)²]dx

Putting the value of dy/dx,

Arc length, L = \(\int_{a}^{b}\)√[1 + (3x1/2)²]dx

L =\(\int_{a}^{b}\)√[1 + 9x]dx

Now, substitute a = 0 and b = 1

L = \(\int_{0}^{1}\)√[1 + 9x]dx

Let, 1 + 9x = p²

Differentiating with respect to x,

9 dx = 2p dp

dx = 2p/9 dp

Since we have substituted the function in terms of p, therefore we have to change the limits.

If x = 0 then,

p = √[1 + 9(0)]

p = 1

If x = 1 then,

p = √[1 + 9(1)]

p = √10

Substitute the value of the limits,

L = \(\int_{1}^{\sqrt{10}}\)p²(2/9) dp

L = 2/9\(\int_{1}^{\sqrt{10}}\)p²dp

L = 2/9 [p³/3] in the limit 1 to √10

L = 2/9 [10√10 - 1](1/3)

L = 2/27 (30.623)

L = 2.268 units

Therefore, the exact length of the curve is 2.268 units.

---

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

Summary:

The exact length of the curve y = 1+2x^3/2, 0 ≤ x ≤ 1 is 2.268 units.