Question

The area of the region bounded by the curve $ y = \max\{|x|, |x-2|\} $, then x-axis and the lines x = -2 and x = 4 is equal to ____.

Hint:

The solution provided in the picture calculates the area as a sum of triangular areas.

Answer 1

Correct Answer: 12

As given in the picture, the area is calculated as:

Required Area = \( \frac{1}{2} \times 2 \times 2 + \frac{1}{2} \times 3 \times 3 + \frac{1}{2} \times 1 \times 11 \) 
Required Area = \( 2 + \frac{9}{2} + \frac{11}{2} \) 
Required Area = \( 2 + \frac{20}{2} \) 
Required Area = \( 2 + 10 \) Required Area = \( 12 \) 
Thus, following the given solution, the area is 12.

Answer 2

To find the area of the region bounded by the curve \( y = \max\{|x|, |x-2|\} \), the x-axis, and the lines \( x = -2 \) and \( x = 4 \), we begin by analyzing the behavior of the function \( y = \max\{|x|, |x-2|\} \).

We consider two key pieces of this piecewise function:
1. \( |x| \): This is a V-shaped function with vertex at the origin (0, 0).
2. \( |x-2| \): This is another V-shaped function with vertex at (2, 0).

Within the interval \( [-2, 4] \), the function \( \max\{|x|, |x-2|\} \) transitions at the point where these two expressions are equal: \( |x| = |x-2| \).

To find this point, we solve the equation:
\( |x| = |x-2| \)
Case 1: \( x \geq 0 \):
\( x = x-2 \Rightarrow 0 = -2 \) (not possible).
\( x = 2-x \Rightarrow 2x = 2 \Rightarrow x = 1 \)
Case 2: \( x < 0 \):
Does not apply because both expressions yield identical values only in \( x \geq 0 \).
Therefore, the transition point is \( x = 1 \).

In intervals:
- When \( x < 1 \), \( \max\{|x|, |x-2|\} = |x-2| \) because for all \( x < 1 \), \( |x-2| \) is greater than or equal to \( |x| \).
- When \( x \geq 1 \), \( \max\{|x|, |x-2|\} = |x| \) as \( |x| \) becomes greater than or equal to \( |x-2| \).

The area calculation consists of integrating these functions over their respective intervals:

\(A = \int_{-2}^{1} |x-2| \, dx + \int_{1}^{4} |x| \, dx \)

Calculating each integral:

1. \( \int_{-2}^{1} |x-2| \, dx \):
\( |x-2| = -(x-2) = 2-x \) for \( x \leq 2 \). Therefore,
\( \int_{-2}^{1} (2-x) \, dx = \left[ 2x - \frac{x^2}{2} \right]_{-2}^{1} = \left( 2(1) - \frac{1}{2} \right) - \left( 2(-2) - \frac{(-2)^2}{2} \right) = \frac{3}{2} - (-3) = \frac{9}{2} \).

2. \( \int_{1}^{4} |x| \, dx \):
\( |x| = x \) for \( x \geq 0 \). Therefore,
\( \int_{1}^{4} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{4} = \frac{4^2}{2} - \frac{1^2}{2} = \frac{16}{2} - \frac{1}{2} = \frac{15}{2} \).

Adding these areas together:
\( A = \frac{9}{2} + \frac{15}{2} = 12 \).

The area of the region is 12, confirming it's within the given range \( [12, 12] \).