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To find the arc length of the function \( y = \frac{2}{3}x^2 - x + 3 \) over the interval \( x = 0 \) to \( x = 2 \), we will use the formula for the arc length of a curve:

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

### Step 1: Differentiate the function
The given function is:
\[
y = \frac{2}{3}x^2 - x + 3
\]

We first need to find the derivative \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{d}{dx} \left( \frac{2}{3}x^2 - x + 3 \right) = \frac{4}{3}x - 1
\]

### Step 2: Square the derivative
Next, we square \( \frac{dy}{dx} \):
\[
\left( \frac{dy}{dx} \right)^2 = \left( \frac{4}{3}x - 1 \right)^2 = \left( \frac{16}{9}x^2 - \frac{8}{3}x + 1 \right)
\]

### Step 3: Set up the arc length integral
The arc length formula becomes:
\[
L = \int_0^2 \sqrt{1 + \left( \frac{16}{9}x^2 - \frac{8}{3}x + 1 \right)} \, dx
\]

Simplify the expression inside the square root:
\[
L = \int_0^2 \sqrt{\frac{16}{9}x^2 - \frac{8}{3}x + 2} \, dx
\]

### Step 4: Calculate the arc length
Let me compute the integral now.The arc length of the curve \( y = \frac{2}{3}x^2 - x + 3 \) over the interval from \( x = 0 \) to \( x = 2 \) is approximately \( 2.557 \) units, rounded to three decimal places.

Would you like a detailed breakdown or have any additional questions?

Here are some related questions to explore:

1. What is the geometric meaning of arc length in calculus?
2. How does changing the interval affect the arc length calculation?
3. Can arc length be calculated for parametric curves?
4. How does the derivative influence the curvature of a function?
5. What techniques can be used to solve more complex arc length integrals?

**Tip**: When solving arc length problems, make sure to simplify the expression inside the square root as much as possible before attempting to compute the integral.

Find the arc length of the graph of the function y = (2/3)x^2 - x + 3 over the interval from x = 0 to x = 2. Round your answer to three decimal places.

Learn how to calculate the arc length of the function y = (2/3)x^2 - x + 3 over the interval [0, 2]. The solution involves taking the derivative, squaring it, and setting up the arc length integral. This calculus problem is suitable for high school seniors or undergraduate students.

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