Question

Let \( D = \{(x, y) \in \mathbb{R}^2 : 0 \leq x \leq 2,\ 0 \leq y \leq 2\} \) and let \( f(t) \) denote the smallest integer greater than or equal to \( t \). Then the value of the integral
\[ \iint_D f(x+y)\, dx\, dy \]
is ______.

Hint:

Split the square by the level lines x+y=1, 2, 3 and weight each strip's area by the ceiling value on it.

Answer 1

Correct Answer: 10

Step 1: Understanding the Question.
Here \( f(t) = \lceil t \rceil \), the ceiling function, the smallest integer that is greater than or equal to \( t \). We need to integrate \( \lceil x+y \rceil \) over the square \( D = [0,2] \times [0,2] \). Since \( x+y \) ranges from \( 0 \) to \( 4 \) on this square, \( \lceil x+y \rceil \) only takes the whole number values \( 1, 2, 3, 4 \) (the single point where \( x+y=0 \) has zero area, so it does not affect the integral).

Step 2: Key Formula or Approach.
Split the square into four regions based on the value of \( s = x+y \): the region where \( 0 < s \leq 1 \) contributes value 1, where \( 1 < s \leq 2 \) contributes value 2, where \( 2 < s \leq 3 \) contributes value 3, and where \( 3 < s \leq 4 \) contributes value 4. The integral becomes
\[ \iint_D \lceil x+y \rceil \, dx\, dy = \sum_{k=1}^{4} k \cdot \text{Area}(k-1 < x+y \leq k) \]
So we just need the area of the square lying between consecutive lines \( x+y=k-1 \) and \( x+y=k \).

Step 3: Compute the area under each line \( x+y \leq s \).
Let \( A(s) \) be the area of the part of the square where \( x+y \leq s \). For \( 0 \leq s \leq 2 \), this region is a right triangle with both legs of length \( s \), so \( A(s) = \frac{s^2}{2} \). For \( 2 \leq s \leq 4 \), the region is the whole square minus a small triangle near the far corner \( (2,2) \) with legs of length \( 4-s \), so \( A(s) = 4 - \frac{(4-s)^2}{2} \). Using these:
\[ A(0)=0,\quad A(1)=\frac{1}{2},\quad A(2)=2,\quad A(3)=4-\frac{1}{2}=\frac{7}{2},\quad A(4)=4 \]

Step 4: Take the differences (the area of each band).
\[ A(1)-A(0) = \frac{1}{2}, \quad A(2)-A(1) = \frac{3}{2}, \quad A(3)-A(2) = \frac{3}{2}, \quad A(4)-A(3) = \frac{1}{2} \]
These four numbers add up to \( 4 \), the total area of the square, which is a good check.

Step 5: Weight each band by its ceiling value and add.
\[ 1\left(\frac{1}{2}\right) + 2\left(\frac{3}{2}\right) + 3\left(\frac{3}{2}\right) + 4\left(\frac{1}{2}\right) = 0.5 + 3 + 4.5 + 2 = 10 \]

Final Answer:
\[ \boxed{10} \]