Question

The area enclosed by \(y = |x-1| + |x-2|\) and \(y = 3\)

Answer 1

Correct Answer: 4

To find the area enclosed by the curves \(y = |x-1| + |x-2|\) and \(y = 3\), we start by understanding the behavior of the given absolute value function. The function \(y = |x-1| + |x-2|\) consists of linear segments due to the absolute values. 
Step 1: Identify the critical points where the expression inside the absolute values change sign, namely at \(x = 1\) and \(x = 2\).
Step 2: Consider the behavior of the function in the intervals defined by these points:

• For \(x < 1\), both expressions \(x - 1\) and \(x - 2\) are negative. Thus, \(y = -(x - 1) - (x - 2) = -x + 1 - x + 2 = -2x + 3\).
• For \(1 ≤ x < 2\), the expression \(x - 1\) is non-negative and \(x - 2\) is negative. Thus, \(y = (x - 1) - (x - 2) = x - 1 - x + 2 = 1.
• For \(x ≥ 2\), both \(x - 1\) and \(x - 2\) are non-negative. Thus, \(y = (x - 1) + (x - 2) = 2x - 3\).

Step 3: Find the intersections with \(y = 3\):

• Solve \(-2x + 3 = 3\):
  \(x = 0\).
• Solve \(1 = 3\): No solution since this is a false statement.
   
• Solve \(2x - 3 = 3\):
  \(2x = 6\), yielding \(x = 3\).

Step 4: The relevant intervals where the curves enclose an area are from \(x = 0\) to \(x = 3\).
Step 5: Integrate the difference between the functions over these intervals:

Interval	Function Difference	Integration
\(x \in [0,1)\)	\(3 - (-2x + 3)\)	\(\int_{0}^{1} (2x) \, dx = [x^2]_{0}^{1} = 1\)
\(x \in [1,2)\)	\(3 - 1\)	\(\int_{1}^{2} (2) \, dx = [2x]_{1}^{2} = 4 - 2 = 2\)
\(x \in [2,3]\)	\(3 - (2x - 3)\)	\(\int_{2}^{3} (6 - 2x) \, dx = [6x - x^2]_{2}^{3} = (18 - 9) - (12 - 4) = 7\)

Step 6: Sum the areas: \(1 + 2 + 1 = 4\), confirming the enclosed area is 4, which lies within the given range \(4,4\).