Home›Calculus 3›Riemann sums for double integrals›Problem 11.466

Riemann sums for double integrals

Problem 11.466 · medium

Estimate \( \displaystyle \iint_R 3 x - y^{2} + 5 \, dA \) over \( \displaystyle R = [0, 4] \times [0, 1] \) with a Riemann sum, \( \displaystyle m = 4 \) subintervals in \( \displaystyle x \) and \( \displaystyle n = 2 \) in \( \displaystyle y \), sampling at the lower left corners.

\[ 1 \cdot \frac{1}{2} \cdot 4 \cdot \frac{1}{4} = \frac{1}{2} \]
ΔA = Δx Δy.✓ Proved
The sample points: (0, 0), (0, 1/2), (1, 0), (1, 1/2), (2, 0), (2, 1/2), (3, 0), (3, 1/2).
Reviewed
\[ 1 \cdot 1 \cdot \frac{1}{2} \left(\frac{19}{4} + 5 + \frac{31}{4} + 8 + \frac{43}{4} + 11 + \frac{55}{4} + 14\right) = \frac{75}{2} \]
ΔA times the sum of f at the sample points.✓ Proved
\[ \int\limits_{0}^{1}\int\limits_{0}^{4} \left(3 x - y^{2} + 5\right)\, dx\, dy = \frac{128}{3} \]
For comparison, the exact value is 128/3.✓ Proved

Answer \( \frac{75}{2} \)

Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
4	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	the sum recomputed in exact fractions

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the grid dimensions, sample points, and area element. The arithmetic in the Riemann sum is correct, and the comparison with the exact integral is accurate.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the grid dimensions, sample points, and area element. The arithmetic in the Riemann sum is correct, and the comparison with the exact integral is accurate.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: fail (error) 2026-10-16 — The sample points listed in step 2 are incorrect for 'lower left corners'. With m=4 and n=2, the grid points are x_i = 0, 1, 2, 3 and y_j = 0, 0.5. The lower-left corner of each subrectangle [x_i, x_{i+1}] x [y_j, y_{j+1}] is (x_i, y_j). The solution lists 8 points, but pairs them incorrectly or implies a different sampling strategy. Specifically, for a fixed x_i, the lower-left corners for the two y-intervals are (x_i, 0) and (x_i, 0.5). The list provided is actually the set of all lower-left corners, which is correct in terms of coordinates, BUT the calculation in step 3 sums f(x,y) for these points. Let's check the values. f(0,0)=5, f(0,0.5)=4.75, f(1,0)=8, f(1,0.5)=7.75, f(2,0)=11, f(2,0.5)=10.75, f(3,0)=14, f(3,0.5)=13.75. Sum = 5+4.75+8+7.75+11+10.75+14+13.75 = 75. Delta A = 1 * 0.5 = 0.5. Sum * Delta A = 37.5 = 75/2. The calculation is correct for the listed points. However, are these the correct points? Lower left corners of the 8 subrectangles. Rectangles are [0,1]x[0,0.5], [1,2]x[0,0.5], [2,3]x[0,0.5], [3,4]x[0,0.5] and same for y in [0.5, 1]. Lower left of [0,1]x[0,0.5] is (0,0). Lower left of [0,1]x[0.5,1] is (0,0.5). Yes, the points are correct. Wait, let me re-read carefully. 'sampling at the lower left corners'. The points listed are indeed the lower-left corners of the 8 subrectangles. The calculation yields 75/2. The exact integral is 128/3 = 42.66. 75/2 = 37.5. This is a valid Riemann sum estimate. Why would it fail? Let's check the prompt again. 'Estimate ... with a Riemann sum'. The solution provides the estimate. Is there an error in the setup? Delta x = (4-0)/4 = 1. Delta y = (1-0)/2 = 0.5. Delta A = 0.5. Points: x in {0,1,2,3}, y in {0, 0.5}. These are the lower-left corners. The sum is correct. The answer is correct. Why did I think it might fail? Maybe the phrasing 'sample points' implies something else? No. Let's look at step 2 again. The points are listed. Step 3 calculates the sum. Step 4 gives the exact value for comparison. Everything seems correct. Is there a subtle error? 'lower left corners'. For the rectangle [3,4]x[0.5,1], the lower left is (3, 0.5). The list includes (3, 0.5). For [3,4]x[0,0.5], lower left is (3,0). List includes (3,0). It seems correct. Let me double check the arithmetic in step 3. 19/4 = 4.75. 5. 31/4 = 7.75. 8. 43/4 = 10.75. 11. 55/4 = 13.75. 14. Sum: 4.75+5=9.75. 7.75+8=15.75. 10.75+11=21.75. 13.75+14=27.75. Total: 9.75+15.75=25.5. 21.75+27.75=49.5. 25.5+49.5=75. 75 * 0.5 = 37.5 = 75/2. The math is right. The points are right. The method is right. Verdict: pass.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the grid dimensions, sample points, and function evaluations. The arithmetic in the Riemann sum is correct, and the comparison with the exact integral is accurate.
gpt-oss:20b: pass 2026-10-15

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/double_riemann, checked 2026-10-18 with SymPy 1.14.0.