Question

Area of the region bounded by the function $f(x)=\begin{cases} x, & x\leq 3 \\ -x+6, & x>3 \end{cases}$ with the $x$-axis (in square units) in the first quadrant is:

• A. \( 18 \)
• B. \( 9 \)
• C. \( 6 \)
• D. \( 3 \)
• E. \( 4.5 \)

Hint:

For piecewise linear graphs, first plot the key points and identify the geometric shape formed. Often, the area can be found much faster using basic geometry than integration.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem asks for the area of a region bounded by a piecewise function and the x-axis in the first quadrant. We can find this area by either sketching the graph and using geometry or by using definite integration. The function defines two line segments.
Step 2: Key Formula or Approach:
Method 1: Geometric Approach
1. Sketch the graph of the function \( f(x) \).
2. Identify the geometric shape of the bounded region.
3. Use the appropriate formula to calculate the area. For a triangle, Area = \( \frac{1}{2} \times \text{base} \times \text{height} \).
Method 2: Integration Approach
The area is given by the definite integral \( \text{Area} = \int_{a}^{b} f(x) dx \). Since the function is piecewise, we split the integral at the point where the definition changes.
Step 3: Detailed Explanation:
Method 1: Geometric Approach
Let's find the key points of the graph.
The function is \( y=x \) for \( x \le 3 \). This is a line passing through (0,0) and (3,3).
The function is \( y=-x+6 \) for \( x>3 \). This line passes through (3,3) (since \( -3+6=3 \)). To find the x-intercept, set \( y=0 \): \( 0 = -x+6 \implies x=6 \). So, it also passes through (6,0).
The region bounded by the function and the x-axis is a triangle with vertices at (0,0), (6,0), and (3,3).
- The base of the triangle lies on the x-axis, from \( x=0 \) to \( x=6 \). So, the base length is \( 6 - 0 = 6 \) units.
- The height of the triangle is the maximum y-value, which occurs at the vertex (3,3). The height is 3 units.
Using the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 3 = 9 \text{ square units} \] Method 2: Integration Approach
We need to integrate from where the function first hits the x-axis (x=0) to where it last hits it (x=6). We split the integral at \( x=3 \):
\[ \text{Area} = \int_0^6 f(x) dx = \int_0^3 x \,dx + \int_3^6 (-x+6) \,dx \] Evaluate the first integral:
\[ \int_0^3 x \,dx = \left[ \frac{x^2}{2} \right]_0^3 = \frac{3^2}{2} - \frac{0^2}{2} = \frac{9}{2} \] Evaluate the second integral:
\[ \int_3^6 (-x+6) \,dx = \left[ -\frac{x^2}{2} + 6x \right]_3^6 = \left( -\frac{6^2}{2} + 6(6) \right) - \left( -\frac{3^2}{2} + 6(3) \right) \] \[ = \left( -18 + 36 \right) - \left( -\frac{9}{2} + 18 \right) = 18 - \left( \frac{-9+36}{2} \right) = 18 - \frac{27}{2} = \frac{36-27}{2} = \frac{9}{2} \] Total Area = \( \frac{9}{2} + \frac{9}{2} = 9 \) square units.
Step 4: Final Answer:
The area of the region is 9 square units. This corresponds to option (B).