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

We will be using the formula of the exact length of the curve to solve this.

Answer: The exact length of the curve y = 4 + 4x^3/2, 0 ≤ x ≤ 1, is 4.149 units.

Let's solve this question step by step.

Explanation:

Given function is y = 4 + 4x^3/2

Now, differentiate the given function (4 + 4x^3/2) with respect to “x”

dy/dx = d(4 + 4x^3/2)/dx

dy/dx = 0 + (3/2) × 4 × x^1/2

dy/dx = 6x^(1/2) ---- (1)

To find arc length, we use the following formula for the length of the arc(L),

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

Putting the value of dy/dx in length of curve formula from (1)

L= \(\int_{a}^{b}\) √[1 + (6x^1/2)²]dx

L= \(\int_{a}^{b}\)√[1 + 36x] dx

Subsitute, a= 0 and b= 1

L=\(\int_{0}^{1}\)√[1 + 36x] dx.

Now, Integrate using substitution method

1+ 36x = u

Differentiate both side w.r.t.x

d(1)/dx + d(36x)/dx = d(u)/dx

dx = du/36

L=\(\int_{0}^{1}\)√[u]/36 du.

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

If x = 0 then, u= 1+36(0) = 1

if x = 1 then, u= 1+36(1) = 37

Substitute the value the limits.

L=\(\int_{1}^{37}\)√[u]/36 du.

⇒ L= (1/36)\(\int_{1}^{37}\)√[u] du

⇒ L= (1/36)\(\int_{1}^{37}\) u^(1/2) du

⇒ L= (1/36)[ u^(3/2) / (3/2) \(]_{1}^{37}\) [Since, \(\int\) xⁿ= (xⁿ⁺¹)/(n +1) ]

⇒ L= (1/36)×(2/3)[ u^(3/2) \(]_{1}^{37}\)

⇒ L= (2/108)[ u^(3/2) \(]_{1}^{37}\)

Put the upper and lower limits.

⇒ L= (2/108)[ 37^(3/2)- 1^(3/2)]

L= 4.149 units.