Question

The area bounded by the curve $y = |x - 2|$, $x = 1$, $x = 3$ and X-axis is

• A. 3 sq. units
• B. 2 sq. units
• C. 1 sq. unit
• D. 4 sq. units

Hint:

When integrating functions involving absolute values, always split the integral at the point where the expression inside the absolute value changes sign. This corresponds to the 'vertex' of the V-shape graph for $|x-a|$.

Answer 1

The Correct Option is A

Step 1: Understand the function $y = |x-2|$.
This function can be written in piecewise form: \[ y = \begin{cases} 2 - x & \text{if } x < 2 \\ x - 2 & \text{if } x \ge 2 \end{cases} \]
Step 2: Set up the integral for the area.
The area is bounded between $x = 1$ and $x = 3$. Since the function changes at $x = 2$, split the integral: \[ \text{Area} = \int_{1}^{3} |x-2| \, dx \] 
Step 3: Split the integral.
\[ \text{Area} = \int_{1}^{2} (2-x)\,dx + \int_{2}^{3} (x-2)\,dx \] 
Step 4: Evaluate the first integral.
\[ \int_{1}^{2} (2-x)\,dx = \left[2x - \frac{x^2}{2}\right]_{1}^{2} \] \[ = \left(4 - 2\right) - \left(2 - \frac{1}{2}\right) = 2 - \frac{3}{2} = \frac{1}{2} \] 
Step 5: Evaluate the second integral.
\[ \int_{2}^{3} (x-2)\,dx = \left[\frac{x^2}{2} - 2x\right]_{2}^{3} \] \[ = \left(\frac{9}{2} - 6\right) - \left(2 - 4\right) = -\frac{3}{2} + 2 = \frac{1}{2} \] 
Step 6: Total area.
\[ \text{Total Area} = \frac{1}{2} + \frac{1}{2} = 1 \] 
Final Answer:
\[ \boxed{1 \text{ sq. unit}} \]