Find the area between y = |x - 1| and y=x from x = - 1 x = 2

To find the area between the curves \( y = |x - 1| \) and \( y = x \) from \( x = -1 \) to \( x = 2 \), we must first break the absolute value function into its piecewise linear components, then integrate the difference between the curves over the given intervals.

### Step 1: Break the absolute value function into two parts
The absolute value function \( y = |x - 1| \) can be written as:

\[
y = 
\begin{cases} 
-(x - 1) = -x + 1 & \text{if } x < 1 \\
x - 1 & \text{if } x \geq 1 
\end{cases}
\]

### Step 2: Identify intersection points of \( y = x \) and \( y = |x - 1| \)
We need to find the points where the curves \( y = |x - 1| \) and \( y = x \) intersect.

- For \( y = -x + 1 \) (when \( x < 1 \)):
  Solve \( x = -x + 1 \):
  \[
  2x = 1 \implies x = \frac{1}{2}
  \]
  
- For \( y = x - 1 \) (when \( x \geq 1 \)):
  Solve \( x = x - 1 \):
  This gives no intersection in this case.

Thus, the curves intersect at \( x = \frac{1}{2} \).

### Step 3: Set up the integrals for the two intervals

We divide the problem into two intervals:
- From \( x = -1 \) to \( x = \frac{1}{2} \), we integrate the difference \( (x - (-x + 1)) = 2x - 1 \).
- From \( x = \frac{1}{2} \) to \( x = 2 \), we integrate the difference \( (x - (x - 1)) = 1 \).

### Step 4: Compute the area

1. **For the interval \( x \in [-1, \frac{1}{2}] \):**
   \[
   \text{Area}_1 = \int_{-1}^{\frac{1}{2}} (2x - 1) \, dx
   \]
   Integrating:
   \[
   \int (2x - 1) \, dx = x^2 - x
   \]
   Evaluate from \( x = -1 \) to \( x = \frac{1}{2} \):
   \[
   \left[ \frac{1}{4} - \frac{1}{2} \right] - \left[ 1 - (-1) \right] = \left( -\frac{1}{4} \right) - (2) = -\frac{9}{4}
   \]

2. **For the interval \( x \in [\frac{1}{2}, 2] \):**
   \[
   \text{Area}_2 = \int_{\frac{1}{2}}^{2} 1 \, dx = \left[ x \right]_{\frac{1}{2}}^{2} = 2 - \frac{1}{2} = \frac{3}{2}
   \]

### Step 5: Total area

The total area is:
\[
\text{Total Area} = \left| -\frac{9}{4} \right| + \frac{3}{2} = \frac{9}{4} + \frac{3}{2} = \frac{9}{4} + \frac{6}{4} = \frac{15}{4}
\]

Thus, the total area between the curves is \( \frac{15}{4} \) square units.

Would you like any details or clarifications on the steps?

Here are some related questions:
1. How does the integration process work for piecewise functions?
2. Can you explain the concept of definite integrals and their geometric interpretation?
3. How would the problem change if we rotated the region around an axis?
4. Can we solve this problem using a different method, such as geometry?
5. How does symmetry play a role in evaluating the area between curves?

**Tip**: When dealing with absolute value functions, always break them into their piecewise linear components for easier integration.

Find the area between y = |x - 1| and y = x from x = -1 to x = 2.

Discover how to calculate the area between the curves y = |x - 1| and y = x from x = -1 to x = 2 using definite integrals. Learn step-by-step how to break down piecewise functions, solve integrals, and apply the Fundamental Theorem of Calculus. Suitable for high school calculus students in Grades 11-12.

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