Problem Analysis #40

A.X-4.0 (72B) · Run 1

24.11s

2,231 toks

Problem Statement

Consider the solid whose base in the $xy$ -plane is the region bounded by the curve

y = \sqrt{x + x \ln x},

the $x$ -axis, and the vertical lines $x = 1$ and $x = 2$ . The cross-sections of this solid taken by planes perpendicular to the $x$ -axis are all equilateral triangles. Find the volume of this solid. [3 points]If the volume is $\frac{\sqrt{3}(p + q\ln 2)}{r}$ , compute $p+q+r$ .\

Ground Truth

Execution Trace

Reasoning Process

[THOUGHT] To find the volume of the solid, we need to integrate the area of the cross-sections along the x-axis from x = 1 to x = 2. The cross-sections are equilateral triangles with their bases lying on the curve $y = \sqrt{x + x \ln x}$ .

Determine the base length of the triangle at each x: The base of the triangle at any point $x$ is $2y = 2\sqrt{x + x \ln x}$ .
Find the area of the equilateral triangle: The formula for the area $A$ of an equilateral triangle with side length $s$ is $A = \frac{\sqrt{3}}{4} s^2$ . Here, the side length $s$ is the same as the base length, $s = 2\sqrt{x + x \ln x}$ . Therefore, the area $A(x)$ is:

A(x) = \frac{\sqrt{3}}{4} \left( 2\sqrt{x + x \ln x} \right)^2 = \frac{\sqrt{3}}{4} \cdot 4(x + x \ln x) = \sqrt{3}(x + x \ln x)

Set up the integral for the volume: The volume $V$ of the solid is given by integrating the area of the cross-sections from $x = 1$ to $x = 2$ :

V = \int_{1}^{2} \sqrt{3}(x + x \ln x) \, dx

Factor out the constant $\sqrt{3}$ :

V = \sqrt{3} \int_{1}^{2} (x + x \ln x) \, dx

Evaluate the integral: We split the integral into two parts:

\int_{1}^{2} (x + x \ln x) \, dx = \int_{1}^{2} x \, dx + \int_{1}^{2} x \ln x \, dx

The first integral is straightforward:

\int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{2^2}{2} - \frac{1^2}{2} = 2 - \frac{1}{2} = \frac{3}{2}

For the second integral, we use integration by parts. Let $u = \ln x$ and $dv = x \, dx$ . Then, $du = \frac{1}{x} \, dx$ and $v = \frac{x^2}{2}$ .

Using integration by parts $\int u \, dv = uv - \int v \, du$ :

\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \int \frac{x}{2} \, dx

= \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx = \frac{x^2}{2} \ln x - \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{2} \ln x - \frac{x^2}{4}

 Evaluate this from $ x = 1 $ to $ x = 2 $:

\left[ \frac{x^2}{2} \ln x - \frac{x^2}{4} \right]_{1}^{2} = \left( \frac{2^2}{2} \ln 2 - \frac{2^2}{4} \right) - \left( \frac{1^2}{2} \ln 1 - \frac{1^2}{4} \right)

= \left( 2 \ln 2 - 1 \right) - \left( 0 - \frac{1}{4} \right) = 2 \ln 2 - 1 + \frac{1}{4} = 2 \ln 2 - \frac{3}{4}

Combine the results:

\int_{1}^{2} (x + x \ln x) \, dx = \frac{3}{2} + 2 \ln 2 - \frac{3}{4} = \frac{6}{4} + 2 \ln 2 - \frac{3}{4} = \frac{3}{4} + 2 \ln 2

Multiply by $\sqrt{3}$ to get the volume:

V = \sqrt{3} \left( \frac{3}{4} + 2 \ln 2 \right) = \frac{\sqrt{3}(3 + 8 \ln 2)}{4}

Therefore, $p = 3$ , $q = 8$ , and $r = 4$ .

Compute $p + q + r$ :

p + q + r = 3 + 8 + 4 = 15

The final answer is:

\boxed{15}

Final Model Output

Comparative Analysis (P40) (Average across runs)

Avg Time Taken

Solar-Open-100B

622.6

K-EXAONE-236B-A23B

191.8

Gemini-3-Pro-Preview

184.9

K-EXAONE-236B-A23B

68.6

Llama-VARCO-8B-Instruct

46.8

Kanana-2-30B-Thinking

40.3

Solar-Pro-2 (31B)(high)

33.8

HCX-007(high)

33.3

Kanana-2-30B-Thinking-2601

31.5

GPT-5.2 (high)

31.0

Deepseek-V3.2

25.5

A.X-4.0 (72B)

24.1

Claude-Opus-4.5

21.5

Grok-4.1-fast

19.4

GPT-5.1 (high)

15.9

EXAONE-4.0.1-32B (high)

3.0

Avg Token Usage

K-EXAONE-236B-A23B

38,155

K-EXAONE-236B-A23B

26,493

Kanana-2-30B-Thinking-2601

8,669

Kanana-2-30B-Thinking

8,605

Solar-Pro-2 (31B)(high)

6,246

GPT-5.2 (high)

6,219

Gemini-3-Pro-Preview

5,568

Solar-Open-100B

5,538

Claude-Opus-4.5

5,101

Grok-4.1-fast

4,946

HCX-007(high)

4,786

Llama-VARCO-8B-Instruct

4,676

Deepseek-V3.2

4,363

GPT-5.1 (high)

3,043

EXAONE-4.0.1-32B (high)

2,876

A.X-4.0 (72B)

2,231