Find the length of the curve y = x^5 / 6 + 1/10x^3 , 1<= x <= 2 , Arc length =

The formula for the length of a curve defined by \( y = f(x) \) from \( x = a \) to \( x = b \) is given by:

\[
L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx
\]

Given \( y = \frac{x^5}{6} + \frac{1}{10x^3} \), we need to first find the derivative \( \frac{dy}{dx} \).

### Step 1: Find \( \frac{dy}{dx} \)

\[
\frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^5}{6} + \frac{1}{10x^3} \right)
\]

Differentiate each term:

\[
\frac{dy}{dx} = \frac{5x^4}{6} - \frac{3}{10x^4}
\]

### Step 2: Set up the arc length formula

The arc length formula becomes:

\[
L = \int_1^2 \sqrt{1 + \left( \frac{5x^4}{6} - \frac{3}{10x^4} \right)^2} \, dx
\]

### Step 3: Simplify and compute

We now need to evaluate this integral numerically. Let me compute this for you.The length of the curve \( y = \frac{x^5}{6} + \frac{1}{10x^3} \) from \( x = 1 \) to \( x = 2 \) is approximately 5.254 units.

Would you like more details on any part of the process? Here are five related questions that could extend your understanding:

1. How do we find the derivative of more complex functions?
2. Can arc length be calculated for parametric curves?
3. How is the arc length formula derived?
4. What role does numerical integration play in finding arc length for complex functions?
5. How does the curve's shape affect its arc length?

**Tip:** For functions that are difficult to integrate exactly, numerical methods like Simpson's rule or using a computer can be helpful.

Calculate the arc length of the curve y = x^5 / 6 + 1 / 10x^3 from x = 1 to x = 2. This problem involves using the arc length formula with differentiation and numerical integration. Step-by-step solution provided for Calculus students.

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