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The given problem is to **evaluate the integral** of \( \int_0^2 (2x - x^2) \, dx \) using the **definition of area as a limit**.

### Step-by-Step Solution:

1. **Definition of Area as a Limit**:
   The integral \( \int_a^b f(x) \, dx \) can be interpreted as the limit of a Riemann sum:
   \[
   \int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x
   \]
   where \( \Delta x = \frac{b - a}{n} \) and \( x_i^* \) is a sample point in each subinterval.

2. **Riemann Sum Setup**:
   For \( \int_0^2 (2x - x^2) \, dx \), we divide the interval \([0, 2]\) into \(n\) equal subintervals. Thus, \( \Delta x = \frac{2 - 0}{n} = \frac{2}{n} \).
   The sample points \( x_i^* \) are chosen as \( x_i^* = \frac{2i}{n} \) for \( i = 1, 2, ..., n \).

3. **Evaluate the Riemann Sum**:
   The function to be integrated is \( f(x) = 2x - x^2 \). Thus, the Riemann sum becomes:
   \[
   \sum_{i=1}^n f(x_i^*) \Delta x = \sum_{i=1}^n \left( 2 \left( \frac{2i}{n} \right) - \left( \frac{2i}{n} \right)^2 \right) \cdot \frac{2}{n}
   \]
   Simplifying the terms inside the sum:
   \[
   = \sum_{i=1}^n \left( \frac{4i}{n} - \frac{4i^2}{n^2} \right) \cdot \frac{2}{n}
   = \frac{2}{n} \sum_{i=1}^n \left( \frac{8i}{n} - \frac{8i^2}{n^2} \right)
   \]

4. **Split the Summation**:
   Now, we can split the sum into two parts:
   \[
   \frac{2}{n} \left( \frac{8}{n} \sum_{i=1}^n i - \frac{8}{n^2} \sum_{i=1}^n i^2 \right)
   \]
   Using the known formulas for the sums of integers and squares of integers:
   \[
   \sum_{i=1}^n i = \frac{n(n+1)}{2}, \quad \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}
   \]
   Substituting these into the expression:
   \[
   \frac{2}{n} \left( \frac{8}{n} \cdot \frac{n(n+1)}{2} - \frac{8}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} \right)
   \]

5. **Simplify and Take the Limit**:
   Simplifying the first term:
   \[
   \frac{2}{n} \cdot \frac{8}{n} \cdot \frac{n(n+1)}{2} = 8 \cdot \left( \frac{n+1}{n} \right)
   \]
   And simplifying the second term:
   \[
   \frac{2}{n} \cdot \frac{8}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{16(n+1)(2n+1)}{3n^2}
   \]
   As \( n \to \infty \), \( \frac{n+1}{n} \to 1 \) and \( \frac{2n+1}{n} \to 2 \). Therefore, the terms become:
   \[
   8 \cdot 1 - \frac{16(1)(2)}{3} = 8 - \frac{32}{3} = \frac{24}{3} - \frac{32}{3} = -\frac{8}{3}
   \]

### Final Answer:
The value of the integral is:
\[
\boxed{\frac{8}{3}}
\]

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Do you have any questions, or would you like further details?

Here are some related questions you might consider:
1. How can you visualize the area under the curve geometrically?
2. How does the choice of sample points in the Riemann sum affect the result?
3. What are other methods of evaluating integrals aside from the limit definition?
4. How does the Fundamental Theorem of Calculus apply to this integral?
5. What is the significance of negative area in definite integrals?

**Tip**: When using the Riemann sum, splitting the summation and using known formulas for sum of integers can greatly simplify the calculation process.

Use the definition of area as a limit to evaluate the integral: ∫(2x - x^2)dx from 0 to 2.

Learn how to evaluate the integral ∫(2x - x^2)dx from 0 to 2 using the definition of area as a limit. This problem covers key calculus concepts, such as Riemann sums and summation formulas. Suitable for advanced high school students or college-level calculus.

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