To solve the problem of finding the area of the region bounded by the function \( f(x) = \max\{x^2, (1-x)^2, 2x(1-x)\} \) for \( 0 \leq x \leq 1 \), we will follow these steps:
### Step 1: Identify the Functions
We have three functions to consider:
1. \( f_1(x) = x^2 \)
2. \( f_2(x) = (1-x)^2 \)
3. \( f_3(x) = 2x(1-x) \)
### Step 2: Find Points of Intersection
To determine which function is the maximum in the interval \( [0, 1] \), we need to find the points of intersection of these functions.
1. **Intersection of \( f_1(x) \) and \( f_2(x) \)**:
\[
x^2 = (1-x)^2
\]
Expanding and rearranging gives:
\[
x^2 = 1 - 2x + x^2 \implies 2x = 1 \implies x = \frac{1}{2}
\]
2. **Intersection of \( f_1(x) \) and \( f_3(x) \)**:
\[
x^2 = 2x(1-x)
\]
Rearranging gives:
\[
x^2 = 2x - 2x^2 \implies 3x^2 - 2x = 0 \implies x(3x - 2) = 0
\]
Thus, \( x = 0 \) or \( x = \frac{2}{3} \).
3. **Intersection of \( f_2(x) \) and \( f_3(x) \)**:
\[
(1-x)^2 = 2x(1-x)
\]
Rearranging gives:
\[
1 - 2x + x^2 = 2x - 2x^2 \implies 3x^2 - 4x + 1 = 0
\]
Using the quadratic formula:
\[
x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6}
\]
Thus, \( x = 1 \) or \( x = \frac{1}{3} \).
### Step 3: Determine the Maximum Function in Each Interval
Now we will analyze the intervals based on the points of intersection \( x = 0, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 1 \).
- **Interval \( [0, \frac{1}{3}] \)**:
- At \( x = 0 \): \( f_1(0) = 0, f_2(0) = 1, f_3(0) = 0 \) → max is \( f_2(x) \).
- **Interval \( [\frac{1}{3}, \frac{1}{2}] \)**:
- At \( x = \frac{1}{3} \): \( f_1(\frac{1}{3}) = \frac{1}{9}, f_2(\frac{1}{3}) = \frac{4}{9}, f_3(\frac{1}{3}) = \frac{4}{9} \) → max is \( f_2(x) \).
- **Interval \( [\frac{1}{2}, \frac{2}{3}] \)**:
- At \( x = \frac{1}{2} \): \( f_1(\frac{1}{2}) = \frac{1}{4}, f_2(\frac{1}{2}) = \frac{1}{4}, f_3(\frac{1}{2}) = \frac{1}{2} \) → max is \( f_3(x) \).
- **Interval \( [\frac{2}{3}, 1] \)**:
- At \( x = \frac{2}{3} \): \( f_1(\frac{2}{3}) = \frac{4}{9}, f_2(\frac{2}{3}) = \frac{1}{9}, f_3(\frac{2}{3}) = \frac{4}{9} \) → max is \( f_1(x) \).
### Step 4: Calculate the Area
Now we will calculate the area under the curve for each segment:
1. **Area from \( 0 \) to \( \frac{1}{3} \)**:
\[
A_1 = \int_0^{\frac{1}{3}} (1-x)^2 \, dx = \left[ x - 2x^2 + \frac{x^3}{3} \right]_0^{\frac{1}{3}} = \left[ \frac{1}{3} - 2 \cdot \frac{1}{9} + \frac{1}{81} \right] = \frac{1}{3} - \frac{2}{9} + \frac{1}{81} = \frac{27 - 18 + 1}{81} = \frac{10}{81}
\]
2. **Area from \( \frac{1}{3} \) to \( \frac{1}{2} \)**:
\[
A_2 = \int_{\frac{1}{3}}^{\frac{1}{2}} (1-x)^2 \, dx = \left[ x - 2x^2 + \frac{x^3}{3} \right]_{\frac{1}{3}}^{\frac{1}{2}} = \left[ \frac{1}{2} - 2 \cdot \frac{1}{4} + \frac{1}{24} \right] - \frac{10}{81}
\]
3. **Area from \( \frac{1}{2} \) to \( \frac{2}{3} \)**:
\[
A_3 = \int_{\frac{1}{2}}^{\frac{2}{3}} 2x(1-x) \, dx = \left[ x^2 - \frac{x^3}{3} \right]_{\frac{1}{2}}^{\frac{2}{3}} = \left[ \frac{4}{9} - \frac{8}{81} \right] - \left[ \frac{1}{4} - \frac{1}{24} \right]
\]
4. **Area from \( \frac{2}{3} \) to \( 1 \)**:
\[
A_4 = \int_{\frac{2}{3}}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{\frac{2}{3}}^{1} = \left[ \frac{1}{3} - \frac{8}{81} \right]
\]
### Step 5: Combine Areas
Finally, we sum all the areas calculated:
\[
\text{Total Area} = A_1 + A_2 + A_3 + A_4
\]
