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This is a Korean CSAT mathematics multiple-choice problem. Solve it and put only the correct choice symbol, one of ①, ②, ③, ④, ⑤, inside \boxed{}.
Problem 26 Consider the solid whose base in thexy-plane is the region bounded by the curve y= √ x+xlnx, thex-axis, and the vertical linesx= 1 andx= 2. The cross-sections of this solid taken by planes perpendicular to thex-axis are all equilateral triangles. Find the volume of this solid. [3 points] ① √ 3(3 + 8 ln 2) 16 ② √ 3(5 + 12 ln 2) 24 ③ √ 3(1 + 12 ln 2) 16 ④ √ 3(1 + 2 ln 2) 4 ⑤ √ 3(1 + 9 ln 2) 12 14
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로그인 회원가입 You are solving a Korean CSAT mathematics problem. Reason carefully, but the final response must contain exactly one final answer inside \boxed{}.
For multiple-choice problems, put only one of ①, ②, ③, ④, ⑤ inside \boxed{}. For short-answer problems, put only the final integer inside \boxed{}. Do not use external tools.
Problem 26 Consider the solid whose base in thexy-plane is the region bounded by the curve y= √ x+xlnx, thex-axis, and the vertical linesx= 1 andx= 2. The cross-sections of this solid taken by planes perpendicular to thex-axis are all equilateral triangles. Find the volume of this solid. [3 points] ① √ 3(3 + 8 ln 2) 16 ② √ 3(5 + 12 ln 2) 24 ③ √ 3(1 + 12 ln 2) 16 ④ √ 3(1 + 2 ln 2) 4 ⑤ √ 3(1 + 9 ln 2) 12 14 더보기
The base of the solid is the region bounded by 𝑦
𝑥 + 𝑥 ln 𝑥 y= x+xlnx
, the 𝑥 x-axis, and the lines 𝑥
1 x=1 and 𝑥
2 x=2. Cross-sections perpendicular to the 𝑥 x-axis are equilateral triangles with side length equal to the 𝑦 y-value of the curve.
The area of an equilateral triangle with side length 𝑠 s is 3 4 𝑠 2 4 3
s 2 . Thus the volume is
𝑉
∫ 1 2 3 4 ( 𝑥 + 𝑥 ln 𝑥 ) 2 𝑑 𝑥
3 4 ∫ 1 2 ( 𝑥 + 𝑥 ln 𝑥 ) 𝑑 𝑥 . V=∫ 1 2
4 3
( x+xlnx
) 2 dx= 4 3
∫ 1 2
(x+xlnx)dx.
Compute the integral:
∫ 1 2 𝑥 𝑑 𝑥
[ 𝑥 2 2 ] 1 2
2 − 1 2
3 2 . ∫ 1 2
xdx=[ 2 x 2
] 1 2
=2− 2 1
= 2 3
.
For ∫ 𝑥 ln 𝑥 𝑑 𝑥 ∫xlnxdx, use integration by parts with 𝑢
ln 𝑥 u=lnx, 𝑑 𝑣
𝑥 𝑑 𝑥 dv=xdx:
∫ 𝑥 ln 𝑥 𝑑 𝑥
𝑥 2 2 ln 𝑥 − ∫ 𝑥 2 2 ⋅ 1 𝑥 𝑑 𝑥
𝑥 2 2 ln 𝑥 − 1 2 ∫ 𝑥 𝑑 𝑥
𝑥 2 2 ln 𝑥 − 𝑥 2 4 . ∫xlnxdx= 2 x 2
lnx−∫ 2 x 2
⋅ x 1
dx= 2 x 2
lnx− 2 1
∫xdx= 2 x 2
lnx− 4 x 2
.
Evaluating from 1 to 2:
[ 𝑥 2 2 ln 𝑥 − 𝑥 2 4 ] 1 2
( 2 ln 2 − 1 ) − ( 0 − 1 4 )
2 ln 2 − 3 4 . [ 2 x 2
lnx− 4 x 2
] 1 2
=(2ln2−1)−(0− 4 1
)=2ln2− 4 3
.
Adding the two parts:
∫ 1 2 ( 𝑥 + 𝑥 ln 𝑥 ) 𝑑 𝑥
3 2 + 2 ln 2 − 3 4
3 4 + 2 ln 2. ∫ 1 2
(x+xlnx)dx= 2 3
+2ln2− 4 3
= 4 3
+2ln2.
Therefore,
𝑉
3 4 ( 3 4 + 2 ln 2 )
3 4 ⋅ 3 + 8 ln 2 4
3 ( 3 + 8 ln 2 ) 16 . V= 4 3
( 4 3
+2ln2)= 4 3
⋅ 4 3+8ln2
= 16 3
(3+8ln2)
.
This matches option ①.
\boxed{①}
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Comparative Analysis (P33) (Average across runs)
Averages use reported measurements only. Missing usage is not treated as zero.