Question

Let the area enclosed by the lines \( x + y = 2 \), \( y = 0 \), \( x = 0 \), and the curve \( f(x) = \min \left\{ x^2 + \frac{3}{4}, 1 + [x] \right\} \), where \( [x] \) denotes the greatest integer less than or equal to \( x \), be \( A \). Then the value of \( 12A \) is ____________.

Hint:

When working with piecewise functions, carefully analyze the behavior of each piece and compute the area step by step for the defined intervals.

Answer 1

Correct Answer: 17

Solution:

1. Understand the boundaries of the region:
   • The region is enclosed by:
     • The line \( x + y = 2 \), rewritten as \( y = 2 - x \),
     • The x-axis (\( y = 0 \)),
     • The y-axis (\( x = 0 \)),
     • The curve \( f(x) = \min \left\{ x^2 + \frac{3}{4}, 1 + [x] \right\} \).
2. Behavior of \( f(x) \):
   • For \( x \in [0, 1) \), \( [x] = 0 \), so \( f(x) = \min \left\{ x^2 + \frac{3}{4}, 1 \right\} \).
     • Here, \( x^2 + \frac{3}{4} \leq 1 \) when \( x^2 \leq \frac{1}{4} \), i.e., \( x \in [0, \frac{1}{2}] \).
     • For \( x \in [\frac{1}{2}, 1) \), \( f(x) = 1 \).
   • For \( x \in [1, 2) \), \( [x] = 1 \), so \( f(x) = \min \left\{ x^2 + \frac{3}{4}, 2 \right\} \).
     • Since \( x^2 + \frac{3}{4} \geq 2 \) for \( x \geq \sqrt{\frac{5}{4}} \), \( f(x) = 2 \).
3. Identify the regions to calculate the area \( A \):
   • For \( x \in [0, \frac{1}{2}] \), \( f(x) = x^2 + \frac{3}{4} \).
   • For \( x \in [\frac{1}{2}, 1] \), \( f(x) = 1 \).
   • For \( x \in [1, 2] \), \( f(x) = 2 \).
4. Set up the integral for the area \( A \):
   • The total area \( A \) is given by:

     \[ A = \int_0^{1/2} \left( 2 - x - \left( x^2 + \frac{3}{4} \right) \right) dx + \int_{1/2}^1 \left( 2 - x - 1 \right) dx + \int_1^2 \left( 2 - x - 2 \right) dx. \]

5. Calculate each integral:
   • For \( x \in [0, \frac{1}{2}] \):

     \[ \int_0^{1/2} \left( 2 - x - x^2 - \frac{3}{4} \right) dx = \int_0^{1/2} \left( \frac{5}{4} - x - x^2 \right) dx. \]

     Solution:

     \[ = \left[ \frac{5}{4}x - \frac{x^2}{2} - \frac{x^3}{3} \right]_0^{1/2} = \frac{5}{8} - \frac{1}{8} - \frac{1}{24} = \frac{5}{12}. \]

   • For \( x \in [\frac{1}{2}, 1] \):

     \[ \int_{1/2}^1 \left( 2 - x - 1 \right) dx = \int_{1/2}^1 \left( 1 - x \right) dx = \left[ x - \frac{x^2}{2} \right]_{1/2}^1. \]

     Solution:

     \[ = \left( 1 - \frac{1}{2} \right) - \left( \frac{1}{2} - \frac{1}{8} \right) = \frac{1}{2} - \frac{3}{8} = \frac{1}{8}. \]

   • For \( x \in [1, 2] \):

     \[ \int_1^2 \left( 2 - x - 2 \right) dx = \int_1^2 \left( -x \right) dx = \left[ -\frac{x^2}{2} \right]_1^2 = -\frac{4}{2} + \frac{1}{2} = -\frac{3}{2}. \]

6. Add up the areas:

   \[ A = \frac{5}{12} + \frac{1}{8} - \frac{3}{2} = \frac{17}{12}. \]

7. Calculate \( 12A \):

   \[ 12A = 12 \cdot \frac{17}{12} = 17. \]

Final Answer: \( \boxed{17} \)