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The problem asks to find the area enclosed by the curves:

\[
y = x^2 \quad \text{and} \quad y = 2x - x^2
\]

### Step 1: Find the Points of Intersection
We need to set the two equations equal to find the points where the curves intersect:

\[
x^2 = 2x - x^2
\]

Rearranging terms:

\[
x^2 + x^2 - 2x = 0
\]

\[
2x^2 - 2x = 0
\]

Factor the equation:

\[
2x(x - 1) = 0
\]

So, the solutions are:

\[
x = 0 \quad \text{or} \quad x = 1
\]

Thus, the curves intersect at \( x = 0 \) and \( x = 1 \).

### Step 2: Set up the Integral for the Area
The area between the curves is given by the integral of the difference between the upper curve and the lower curve. Here, the curve \( y = 2x - x^2 \) is above \( y = x^2 \) in the interval from \( x = 0 \) to \( x = 1 \).

The area is:

\[
A = \int_0^1 \left[ (2x - x^2) - x^2 \right] \, dx
\]

Simplifying the integrand:

\[
A = \int_0^1 (2x - 2x^2) \, dx
\]

### Step 3: Compute the Integral
Now, let's compute the integral:

\[
A = \int_0^1 2x \, dx - \int_0^1 2x^2 \, dx
\]

For the first integral:

\[
\int_0^1 2x \, dx = \left[ x^2 \right]_0^1 = 1^2 - 0^2 = 1
\]

For the second integral:

\[
\int_0^1 2x^2 \, dx = \left[ \frac{2x^3}{3} \right]_0^1 = \frac{2(1^3)}{3} - \frac{2(0^3)}{3} = \frac{2}{3}
\]

### Step 4: Calculate the Area
Now, subtract the two results:

\[
A = 1 - \frac{2}{3} = \frac{1}{3}
\]

### Final Answer:
The area enclosed by the curves is \( \frac{1}{3} \).

---

Would you like further details or clarification on any part of this solution?

Here are 5 related questions to explore:

1. How do you find the points of intersection between any two curves?
2. What is the general method for setting up an area integral between two curves?
3. How would the area calculation change if one of the curves was below the x-axis?
4. What is the significance of the limits of integration in problems like this?
5. How do you interpret the result geometrically on the coordinate plane?

**Tip:** When setting up integrals for areas between curves, always identify which curve is on top and which is on the bottom over the given interval.

Encuentra el área encerrada por las curvas y = x^2 y y = 2x - x^2

Learn how to calculate the area enclosed by the curves y = x^2 and y = 2x - x^2 using definite integrals. This step-by-step solution explains the process of finding the points of intersection and setting up the integral for the area. Suitable for advanced calculus students.

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