from a handpicked tutor in LIVE 1-to-1 classes

# 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, \(\int\) x^{n }= (x^{n+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.

### Hence, the exact length of the curve y = 4 + 4x^{3/2}, 0 ≤ x ≤ 1, is 4.149 units.

visual curriculum