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To find the arc length of the function \( f(x) = \frac{x^2}{2} \) from \( x = 0 \) to \( x = \sqrt{3} \), we'll use the arc length formula for a function \( y = f(x) \):

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

Given:
- \( f(x) = \frac{x^2}{2} \)
- Limits of integration: \( a = 0 \) and \( b = \sqrt{3} \)

### Step 1: Compute the derivative of \( f(x) \)
\[
\frac{dy}{dx} = f'(x) = x
\]

### Step 2: Substitute the derivative into the arc length formula
\[
L = \int_{0}^{\sqrt{3}} \sqrt{1 + (x)^2} \, dx
\]

### Step 3: Simplify the integral expression
\[
L = \int_{0}^{\sqrt{3}} \sqrt{1 + x^2} \, dx
\]

### Step 4: Solve the integral
To integrate \( \int \sqrt{1 + x^2} \, dx \), we can use a trigonometric substitution, such as \( x = \sinh(u) \), or a hyperbolic trigonometric approach. However, a more straightforward way might be using a standard integral formula. 

Let's use the standard integral:
\[
\int \sqrt{1 + x^2} \, dx = \frac{1}{2} \left( x \sqrt{1 + x^2} + \sinh^{-1}(x) \right)
\]

### Step 5: Evaluate the definite integral from 0 to \( \sqrt{3} \)
Substitute \( x = \sqrt{3} \) and \( x = 0 \) into the formula:
\[
L = \frac{1}{2} \left[ \sqrt{3} \cdot \sqrt{1 + (\sqrt{3})^2} + \sinh^{-1}(\sqrt{3}) \right] - \frac{1}{2} \left[ 0 \cdot \sqrt{1 + 0^2} + \sinh^{-1}(0) \right]
\]

Calculating these terms:
- \( \sqrt{1 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \)
- \( \sinh^{-1}(\sqrt{3}) = \ln(\sqrt{3} + 2) \)
- \( \sinh^{-1}(0) = 0 \)

So the expression becomes:
\[
L = \frac{1}{2} \left[ \sqrt{3} \cdot 2 + \ln(\sqrt{3} + 2) \right] = \frac{1}{2} \left[ 2\sqrt{3} + \ln(\sqrt{3} + 2) \right]
\]

\[
L = \sqrt{3} + \frac{1}{2} \ln(\sqrt{3} + 2)
\]

Therefore, the arc length of the function \( f(x) = \frac{x^2}{2} \) from \( x = 0 \) to \( x = \sqrt{3} \) is:
\[
L = \sqrt{3} + \frac{1}{2} \ln(\sqrt{3} + 2)
\]

Would you like more details on this solution or have any questions?

### Related Questions:
1. What is the geometric meaning of the arc length of a curve?
2. How can trigonometric substitutions simplify the process of finding integrals?
3. What other methods can be used to solve the integral \( \int \sqrt{1 + x^2} \, dx \)?
4. How do we interpret the result of the arc length in real-life applications?
5. What is the inverse hyperbolic sine function and how is it derived?

### Tip:
When dealing with arc length integrals involving square roots, substitutions like trigonometric or hyperbolic functions often simplify the integration process significantly.

Find the arc length of f(x) = x^2/2 from x = 0 to x = sqrt(3).

Learn how to compute the arc length of the function f(x) = x^2/2 over the interval from x = 0 to x = sqrt(3). This solution involves key calculus concepts such as derivatives, integrals, and the arc length formula. Suitable for students studying calculus.

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