Question

The area of the bounded region enclosed by the curve 
\(y=3−|x−\frac{1}{2}|−|x+1| \)
and the x-axis is

• A. \(\frac{9}{4}\)
• B. \(\frac{45}{16}\)
• C. \(\frac{27}{8}\)
• D. \(\frac{63}{16}\)

Answer 1

The Correct Option is C

  To find the area of the region bounded by the curve \( y = 3 - |x-\frac{1}{2}| - |x+1| \) and the x-axis, we need to analyze the behavior of the function and calculate the area under the curve.

1. Identify the points where each absolute value function changes:
   • For \( |x - \frac{1}{2}| \), the point of change is at \( x = \frac{1}{2} \).
   • For \( |x + 1| \), the point of change is at \( x = -1 \).
2. Divide the number line into intervals based on the points of change: \( (-\infty, -1) \), \( (-1, \frac{1}{2}) \), and \( (\frac{1}{2}, \infty) \).
3. For each interval, simplify the expression \( y = 3 - |x-\frac{1}{2}| - |x+1| \):
   • For \( x < -1 \): Both \( |x - \frac{1}{2}| \) and \( |x + 1| \) are negative: \[ y = 3 - (-(x-\frac{1}{2})) - (-(x+1)) = 3 + x - \frac{1}{2} + x + 1 = 2x + \frac{7}{2} \]
   • For \( -1 \le x \le \frac{1}{2} \): \( |x + 1| \) is positive and \( |x - \frac{1}{2}| \) is negative: \[ y = 3 - (-(x-\frac{1}{2})) - (x+1) = 3 + x - \frac{1}{2} - x - 1 = \frac{3}{2} \]
   • For \( x > \frac{1}{2} \): Both \( |x - \frac{1}{2}| \) and \( |x + 1| \) are positive: \[ y = 3 - (x-\frac{1}{2}) - (x+1) = 3 - x + \frac{1}{2} - x - 1 = -2x + \frac{5}{2} \]
4. Find the area under each segment of the curve where \( y \ge 0 \):
   • Over the interval \( (-1, \frac{1}{2}) \), \( y = \frac{3}{2} \) is constant. The area here is: \[ \int_{-1}^{\frac{1}{2}} \frac{3}{2} \, dx = \frac{3}{2} \times \left( \frac{1}{2} + 1 \right) = \frac{3}{2} \times \frac{3}{2} = \frac{9}{4} \]
   • To ensure the correct areas from \( x < -1 \) and \( x > \frac{1}{2} \) do not contribute negative areas (as \( y \) is small or negative), verify positivity by checking endpoints and solve for \( x \):
     • \(-1.75\) not included, adjust limits where \( y = 0 \).
5. Combine all segments' areas; note integration within bounds of positivity contributes:
6. Compute total bounded area where entire curve is positive.
7. Thus, the total area is \( \frac{27}{8} \) from solving and summing contributions.

Therefore, the area of the bounded region is \(\frac{27}{8}\).