Question

The area of the region bounded by the curves \( y = |x - 2| \), \( x = 1 \), \( x = 3 \) and \( y = 0 \) is:

• A. \( 4 \)
• B. \( 12 \)
• C. \( 3 \)
• D. \( 14 \)
• E. \( 1 \)

Hint:

For modulus graphs like \(y = |x-a|\), always split the interval at \(x=a\). This prevents incorrect integration and makes the graph easy to visualize.

Answer 1

The Correct Option is C

Concept: For modulus functions: \[ |x-a| = \begin{cases} a-x, & x \\ x-a, & x \ge a \end{cases} \] So we must split the interval at the point where the modulus changes sign.

Step 1: Break the modulus function. Given: \[ y = |x - 2| \] So, \[ y = \begin{cases} 2-x, & x \\ x-2, & x \ge 2 \end{cases} \] Since limits are from \[ x = 1 \text{to} x = 3 \] we split at \[ x = 2 \]

Step 2: Find area from \(x=1\) to \(x=2\). \[ A_1 = \int_{1}^{2} (2-x)\,dx \] \[ = \left[ 2x - \frac{x^2}{2} \right]_{1}^{2} \] \[ = \left(4 - 2\right) - \left(2 - \frac{1}{2}\right) \] \[ = 2 - \frac{3}{2} = \frac{1}{2} \]

Step 3: Find area from \(x=2\) to \(x=3\). \[ A_2 = \int_{2}^{3} (x-2)\,dx \] \[ = \left[ \frac{x^2}{2} - 2x \right]_{2}^{3} \] \[ = \left( \frac{9}{2} - 6 \right) - \left( 2 - 4 \right) \] \[ = -\frac{3}{2} + 2 = \frac{1}{2} \]

Step 4: Total area \[ A = A_1 + A_2 \] \[ A = \frac{1}{2} + \frac{1}{2} \] \[ A = 1 \] Hence, \[ \boxed{1} \]