Problem Analysis #calculus_24

Motif 3 · Run 1

Python ● execution succeeded
Correct
6.74s
1,976 toks

Infron · Motif 3 · 46문항 × 3회 · Python TAR. 131,072 output tokens · 기본 1,800초 · 6+1 turns. ACC는 전체 138회, Pass@1은 각 문항의 첫 유효 시도, Pass@3는 3회 내 정답으로 계산합니다.

Problem Statement

∫0π/2sin⁡x−sin⁡3x dx\int_0^{\pi/2}\sqrt{\sin x-\sin^3x}\,dx의 값은?

① 16\frac16 ② 13\frac13 ③ 12\frac12 ④ 23\frac23 ⑤ 56\frac56

최종 답은 선택지 번호 1~5 중 하나를 \boxed{} 안에 쓰세요.

Ground Truth
4
Recorded model response / solution
Reasoning Process

Recorded Python call:

import mpmath as mp
f = lambda x: mp.sqrt(mp.sin(x) - mp.sin(x)**3)
print(mp.quad(f, [0, mp.pi/2]))

Tool Output
Execution succeeded 0.666666666666667
Reasoning Process
∫0π/2sin⁡x−sin⁡3x dx=∫0π/2sin⁡xcos⁡2x dx=∫0π/2cos⁡xsin⁡x dx.\int_0^{\pi/2}\sqrt{\sin x-\sin^3x}\,dx = \int_0^{\pi/2}\sqrt{\sin x\cos^2x}\,dx = \int_0^{\pi/2}\cos x\sqrt{\sin x}\,dx.

Let u=sin⁡xu=\sin x, so du=cos⁡x dxdu=\cos x\,dx. Then

∫01u du=[23u3/2]01=23.\int_0^1 \sqrt{u}\,du = \left[\frac{2}{3}u^{3/2}\right]_0^1 = \frac{2}{3}.

따라서 정답은 ④입니다.

4\boxed{4}
Final Model Output
4

Comparative Analysis (Pcalculus_24) (Average across runs)

Averages use reported measurements only. Missing usage is not treated as zero.

Avg Time Taken

Solar Pro 4
35.5
MiMo V2.6 Pro
32.1
Union Alpha
28.6
Grok 4.7
16.2
Motif 3
14.9
Muse Spark 1.3
10.3
DeepSeek V4.1 Flash
4.9
Solar Mini 4 Preview
3.8

Avg Token Usage

Grok 4.7
7,674
Solar Pro 4
7,009
Motif 3
5,636
MiMo V2.6 Pro
4,422
Muse Spark 1.3
4,236
Union Alpha
4,000
Solar Mini 4 Preview
2,598
DeepSeek V4.1 Flash
2,508
    CSAT 2026 · Problem calculus_24 · Motif 3 | EntropyMath