To find the area of the region bounded by the curves \( y = (x-1)^2 \) and \( y = |x-1| \), we will follow these steps:
### Step 1: Identify the curves
1. The first curve is a parabola given by \( y = (x-1)^2 \).
2. The second curve is an absolute value function given by \( y = |x-1| \).
### Step 2: Find the points of intersection
To find the area between the curves, we first need to find their points of intersection. We set the equations equal to each other:
\[
(x-1)^2 = |x-1|
\]
We can break this into two cases based on the definition of the absolute value.
**Case 1:** \( x - 1 \geq 0 \) (i.e., \( x \geq 1 \))
In this case, \( |x-1| = x - 1 \). Thus, we have:
\[
(x-1)^2 = x - 1
\]
Rearranging gives:
\[
(x-1)^2 - (x-1) = 0
\]
\[
(x-1)(x-2) = 0
\]
This gives us \( x = 1 \) and \( x = 2 \).
**Case 2:** \( x - 1 < 0 \) (i.e., \( x < 1 \))
In this case, \( |x-1| = 1 - x \). Thus, we have:
\[
(x-1)^2 = 1 - x
\]
Rearranging gives:
\[
(x-1)^2 + x - 1 = 0
\]
\[
x^2 - 2x + 1 + x - 1 = 0
\]
\[
x^2 - x = 0
\]
\[
x(x-1) = 0
\]
This gives us \( x = 0 \) and \( x = 1 \).
### Step 3: Identify the points of intersection
The points of intersection are \( (1, 0) \) and \( (2, 1) \).
### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = 1 \) to \( x = 2 \) can be calculated using the integral:
\[
A = \int_{1}^{2} \left( |x-1| - (x-1)^2 \right) \, dx
\]
Since \( x \geq 1 \) in this interval, we have \( |x-1| = x-1 \). Thus, the integral becomes:
\[
A = \int_{1}^{2} \left( (x-1) - (x-1)^2 \right) \, dx
\]
### Step 5: Simplify the integrand
Now, simplify the integrand:
\[
A = \int_{1}^{2} \left( (x-1) - (x^2 - 2x + 1) \right) \, dx
\]
\[
= \int_{1}^{2} \left( x - 1 - x^2 + 2x - 1 \right) \, dx
\]
\[
= \int_{1}^{2} \left( -x^2 + 3x - 2 \right) \, dx
\]
### Step 6: Integrate
Now we can integrate:
\[
A = \left[ -\frac{x^3}{3} + \frac{3x^2}{2} - 2x \right]_{1}^{2}
\]
Calculating at the bounds:
1. At \( x = 2 \):
\[
-\frac{2^3}{3} + \frac{3(2^2)}{2} - 2(2) = -\frac{8}{3} + 6 - 4 = -\frac{8}{3} + 2 = -\frac{8}{3} + \frac{6}{3} = -\frac{2}{3}
\]
2. At \( x = 1 \):
\[
-\frac{1^3}{3} + \frac{3(1^2)}{2} - 2(1) = -\frac{1}{3} + \frac{3}{2} - 2 = -\frac{1}{3} + \frac{3}{2} - \frac{6}{3} = -\frac{1}{3} + \frac{9}{6} - \frac{6}{3} = -\frac{1}{3} + \frac{3}{6} = -\frac{1}{3} + \frac{1}{2}
\]
Calculating gives:
\[
-\frac{1}{3} + \frac{3}{6} = -\frac{1}{3} + \frac{1}{2} = \frac{-2 + 3}{6} = \frac{1}{6}
\]
### Step 7: Calculate the area
Now substituting back into the area calculation:
\[
A = \left( -\frac{2}{3} - \frac{1}{6} \right) = -\frac{4}{6} - \frac{1}{6} = -\frac{5}{6}
\]
Taking the absolute value gives:
\[
A = \frac{5}{6}
\]
Since we have two symmetrical areas, the total area is:
\[
\text{Total Area} = 2 \times \frac{5}{6} = \frac{5}{3}
\]
### Final Answer
Thus, the area of the region bounded by the curves is:
\[
\frac{5}{3} \text{ square units.}
\]
