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: Understand the piecewise function.
The function is given as \[ f(x)= \begin{cases} x, & x\leq 3 -x+6, & x>3 \end{cases} \] So for \[ x\leq 3 \] the graph is the straight line \[ y=x \] and for \[ x>3 \] the graph is the straight line \[ y=-x+6 \]

Step 2: Find where the graph meets the \( x \)-axis in the first quadrant.
For the first part, \[ y=x \] meets the \( x \)-axis at \[ x=0 \] So one point is \[ (0,0) \] For the second part, \[ y=-x+6 \] meets the \( x \)-axis when \[ -x+6=0 \] \[ x=6 \] So the other intercept is \[ (6,0) \]

Step 3: Find the joining point of the two lines.
At \[ x=3 \] for the first branch, \[ y=3 \] and for the second branch, \[ y=-3+6=3 \] So both branches meet at \[ (3,3) \] Thus the graph forms a triangular shape in the first quadrant with vertices \[ (0,0),\ (3,3),\ (6,0) \]

Step 4: Identify the bounded region.
The bounded region with the \( x \)-axis is the triangle formed by: \[ (0,0),\ (3,3),\ (6,0) \] Its base lies on the \( x \)-axis from \[ x=0 \text{ to } x=6 \] So the base length is \[ 6 \] and the height is the vertical distance from \[ (3,3) \] to the \( x \)-axis, which is \[ 3 \]

Step 5: Use the triangle area formula.
Area of triangle \[ =\frac{1}{2}\times \text{base}\times \text{height} \] So, \[ \text{Area}=\frac{1}{2}\times 6\times 3 \] \[ =3\times 3=9 \]

Step 6: Verify by integration as a check.
We can also compute the area as \[ \int_0^3 x\,dx+\int_3^6 (-x+6)\,dx \] Now, \[ \int_0^3 x\,dx=\left[\frac{x^2}{2}\right]_0^3=\frac{9}{2} \] and \[ \int_3^6 (-x+6)\,dx=\left[-\frac{x^2}{2}+6x\right]_3^6 \] \[ =\left(-18+36\right)-\left(-\frac{9}{2}+18\right) \] \[ =18-\frac{27}{2}=\frac{9}{2} \] Total area: \[ \frac{9}{2}+\frac{9}{2}=9 \] So the result is confirmed.

Step 7: Final conclusion.
Hence, the required area is \[ \boxed{9} \] Therefore, the correct option is \[ \boxed{(2)\ 9} \]