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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 28 Consider the function f(x) = 1 2 x2 −x+ ln(1 +x), and lettbe a positive real number. Fors >0, let (s, f(s)) be a point on the curve y=f(x). Letg(t) denote the value ofssuch that the distance between the foot of the perpendicular from (s, f(s)) to they-axis and the point where the tangent to the curve at (s, f(s)) meets they-axis is equal tot. Evaluate Z 27 4 1 2 g(t)dt. [4 points] ① 161 12 + ln 3 ② 40 3 + ln 3 ③ 53 4 + ln 2 ④ 79 6 + ln 2 ⑤ 157 12 + ln 2 15 Numerical answer 더보기

Looking at this problem, I need to find f'(x), set up the distance condition, express g(t), and evaluate the integral.

Step 1: Find f'(x)

𝑓 ( 𝑥 )

1 2 𝑥 2 − 𝑥 + ln ⁡ ( 1 + 𝑥 ) f(x)= 2 1 ​

x 2 −x+ln(1+x)

𝑓 ′ ( 𝑥 )

𝑥 − 1 + 1 1 + 𝑥

𝑥 ( 1 + 𝑥 ) − ( 1 + 𝑥 ) + 1 1 + 𝑥

𝑥 2 1 + 𝑥 f ′ (x)=x−1+ 1+x 1 ​

= 1+x x(1+x)−(1+x)+1 ​

= 1+x x 2 ​

Step 2: Set up the distance

Foot of perpendicular from (s, f(s)) to y-axis: (0, f(s)) Tangent at (s, f(s)): y - f(s) = f'(s)(x - s); at x=0: y = f(s) - s·f'(s) Distance = |f(s) - (f(s) - s·f'(s))| = s·f'(s) = s · s²/(1+s) = s³/(1+s)

So t = s³/(1+s), and g(t) = s.

Step 3: Change of variables With t = s³/(1+s):

𝑑 𝑡 𝑑 𝑠

3 𝑠 2 ( 1 + 𝑠 ) − 𝑠 3 ( 1 + 𝑠 ) 2

𝑠 2 ( 3 + 2 𝑠 ) ( 1 + 𝑠 ) 2 ds dt ​

= (1+s) 2 3s 2 (1+s)−s 3 ​

= (1+s) 2 s 2 (3+2s) ​

The integral becomes:

∫ 4 27 1 2 𝑔 ( 𝑡 )   𝑑 𝑡

1 2 ∫ 𝑠 1 𝑠 2 𝑠 ⋅ 𝑠 2 ( 3 + 2 𝑠 ) ( 1 + 𝑠 ) 2   𝑑 𝑠

1 2 ∫ 𝑠 1 𝑠 2 𝑠 3 ( 3 + 2 𝑠 ) ( 1 + 𝑠 ) 2   𝑑 𝑠 ∫ 4 27 ​

2 1 ​

g(t)dt= 2 1 ​

∫ s 1 ​

s 2 ​

​

s⋅ (1+s) 2 s 2 (3+2s) ​

ds= 2 1 ​

∫ s 1 ​

s 2 ​

​

(1+s) 2 s 3 (3+2s) ​

ds

Step 4: Polynomial division

2 𝑠 4 + 3 𝑠 3 ( 𝑠 + 1 ) 2

2 𝑠 2 − 𝑠 + 𝑠 ( 𝑠 + 1 ) 2 (s+1) 2 2s 4 +3s 3 ​

=2s 2 −s+ (s+1) 2 s ​

Step 5: Integrate

1 2 ∫ ( 2 𝑠 2 − 𝑠 + 𝑠 ( 𝑠 + 1 ) 2 ) 𝑑 𝑠

𝑠 3 3 − 𝑠 2 4 + 1 2 ln ⁡ ( 𝑠 + 1 ) + 1 2 ( 𝑠 + 1 ) + 𝐶 2 1 ​

∫(2s 2 −s+ (s+1) 2 s ​

)ds= 3 s 3 ​

− 4 s 2 ​

2 1 ​

ln(s+1)+ 2(s+1) 1 ​

+C

Step 6: Determine limits

When t = 4: s³/(1+s) = 4 → s = 2 works: 8/3 ≠ 4. Let me check: 2³/(1+2) = 8/3. Not 4.

Let me reconsider — checking s = 2: 8/3 ≈ 2.67. Checking s = 3: 27/4 = 6.75. So between 2 and 3.

Actually, let me check s = 2 for t = 4: 2³/(1+2) = 8/3. Not 4.

Hmm, let me reconsider. Perhaps I should check if there's a cleaner relationship.

Actually, let me reconsider the limits. For t = 27: s³/(1+s) = 27. If s = 3: 27/4 = 6.75. If s = 6: 216/7 ≈ 30.86. So between 5 and 6.

These don't give clean values. Let me reconsider the problem setup.

Wait — let me reconsider. Maybe the limits correspond to specific s values. Let me check s = 2 and s = 3:

s = 2: t = 8/3 s = 3: t = 27/4

Neither gives 4 or 27 directly. Let me reconsider.

Actually, let me reconsider whether the integral limits might correspond to s = 2 and s = 3, giving t = 8/3 and t = 27/4. That doesn't match 4 and 27 either.

Let me try a different approach. Let me check if s = 2 gives t = 4: 2³/(1+2) = 8/3 ≠ 4.

Hmm, let me reconsider. Maybe I need to check s values more carefully.

For t = 4: s³ = 4(1+s) = 4 + 4s → s³ - 4s - 4 = 0 For t = 27: s³ = 27(1+s) = 27 + 27s → s³ - 27s - 27 = 0

These don't have nice rational roots. Let me reconsider the problem.

Actually, wait. Let me reconsider. Maybe I should check if the integral evaluates to one of the answer choices regardless of the exact s values, using the relationship between s and t.

Let me compute the definite integral using the antiderivative F(s) = s³/3 - s²/4 + (1/2)ln(s+1) + 1/(2(s+1)):

The integral = F(s₂) - F(s₁)

where s₁ satisfies s₁³/(1+s₁) = 4 and s₂ satisfies s₂³/(1+s₂) = 27.

Let me try to express F(s) in terms of t. Since t = s³/(1+s), we have s³ = t(1+s).

s³/3 = t(1+s)/3

So F(s) = t(1+s)/3 - s²/4 + (1/2)ln(s+1) + 1/(2(s+1))

This still depends on s, not just t. So the integral depends on the specific s values.

Let me try to numerically estimate. For t = 4: s³ - 4s - 4 = 0. s ≈ 2.38: 13.485 - 9.52 - 4 = -0.035 s ≈ 2.382: 13.52 - 9.528 - 4 = -0.008 s ≈ 2.383: 13.54 - 9.532 - 4 = 0.008 So s₁ ≈ 2.382

For t = 27: s³ - 27s - 27 = 0. s ≈ 5.7: 185.193 - 153.9 - 27 = 4.293 s ≈ 5.6: 175.616 - 151.2 - 27 = -2.584 s ≈ 5.65: 180.36 - 152.55 - 27 = 0.81 s ≈ 5.63: 178.45 - 152.01 - 27 = -0.56 s ≈ 5.64: 179.4 - 152.28 - 27 = 0.12 So s₂ ≈ 5.639

Now F(5.639) - F(2.382): F(5.639) = 5.639³/3 - 5.639²/4 + (1/2)ln(6.639) + 1/(2·6.639) = 179.4/3 - 31.8/4 + 0.5·1.893 + 1/13.278 = 59.8 - 7.95 + 0.947 + 0.0753 = 52.87

F(2.382) = 2.382³/3 - 2.382²/4 + (1/2)ln(3.382) + 1/(2·3.382) = 13.52/3 - 5.674/4 + 0.5·1.218 + 1/6.764 = 4.507 - 1.419 + 0.609 + 0.148 = 3.845

Difference ≈ 52.87 - 3.845 = 49.03

Hmm, let me check the answer choices: ① 161/12 + ln 3 ≈ 13.42 + 1.099 = 14.52 ② 40/3 + ln 3 ≈ 13.33 + 1.099 = 14.43 ③ 53/4 + ln 2 ≈ 13.25 + 0.693 = 13.94 ④ 79/6 + ln 2 ≈ 13.17 + 0.693 = 13.86 ⑤ 157/12 + ln 2 ≈ 13.08 + 0.693 = 13.77

None of these match 49.03. I must be making an error somewhere.

Let me reconsider. Maybe the integral is not ∫(1/2)g(t)dt but something else. Let me re-read: "Z 27 4 1 2 g(t)dt" — this seems to be ∫[4 to 27] (1/2)g(t)dt.

Hmm, but my numerical answer doesn't match any option. Let me reconsider the problem.

Wait, maybe I should reconsider the distance. Let me re-read: "the distance between the foot of the perpendicular from (s, f(s)) to the y-axis and the point where the tangent to the curve at (s, f(s)) meets the y-axis."

Foot of perpendicular from (s, f(s)) to y-axis: The y-axis is x = 0. The perpendicular from (s, f(s)) to the y-axis is horizontal, so the foot is (0, f(s)). ✓

Tangent meets y-axis at (0, f(s) - s·f'(s)). ✓

Distance = |s·f'(s)| = s³/(1+s). ✓

Hmm, let me reconsider. Maybe the problem is asking for a different integral or I have the wrong f(x).

f(x) = 1/(2x²) - x + ln(1+x)? No, it says 1/2 x² - x + ln(1+x).

Wait, let me re-read: "f(x) = 1 2 x2 −x+ ln(1 +x)"

This could be f(x) = 1/(2x²) - x + ln(1+x) or f(x) = (1/2)x² - x + ln(1+x).

If it's f(x) = 1/(2x²) - x + ln(1+x), then f'(x) = -1/x³ - 1 + 1/(1+x). That seems messy.

Let me try f(x) = (1/2)x² - x + ln(1+x) again, which gave f'(x) = x²/(1+x).

Actually, wait. Let me reconsider. Maybe I should check if the limits 4 and 27 correspond to s = 2 and s = 3 in a different way.

If s = 2: t = 8/3. If s = 3: t = 27/4.

Hmm, 8/3 and 27/4... not 4 and 27.

But what if the integral is from t = 8/3 to t = 27/4? Then s goes from 2 to 3.

Let me compute F(3) - F(2): F(3) = 27/3 - 9/4 + (1/2)ln(4) + 1/8 = 9 - 2.25 + ln(2) + 0.125 = 6.875 + ln(2) F(2) = 8/3 - 1 + (1/2)ln(3) + 1/6 = 8/3 - 1 + 1/6 + (1/2)ln(3) = 8/3 - 6/6 + 1/6 + (1/2)ln(3) = 8/3 - 5/6 + (1/2)ln(3) = 16/6 - 5/6 + (1/2)ln(3) = 11/6 + (1/2)ln(3)

F(3) - F(2) = 6.875 + ln(2) - 11/6 - (1/2)ln(3) = 55/8 - 11/6 + ln(2) - (1/2)ln(3) = (165 - 44)/24 + ln(2) - (1/2)ln(3) = 121/24 + ln(2) - (1/2)ln(3)

That doesn't match either.

Hmm, let me reconsider. Maybe the limits are t = 4 and t = 27, and I need to find the corresponding s values more carefully, or maybe I made an error in the antiderivative.

Actually, let me reconsider. Let me recheck my polynomial division.

(2s⁴ + 3s³)/(s+1)²:

2s⁴ + 3s³ divided by s² + 2s + 1:

2s²(s² + 2s + 1) = 2s⁴ + 4s³ + 2s² Remainder: (2s⁴ + 3s³) - (2s⁴ + 4s³ + 2s²) = -s³ - 2s²

-s(s² + 2s + 1) = -s³ - 2s² - s Remainder: (-s³ - 2s²) - (-s³ - 2s² - s) = s

So (2s⁴ + 3s³)/(s+1)² = 2s² - s + s/(s+1)² ✓

∫(2s² - s)ds = 2s³/3 - s²/2

Wait! I think I made an error. Let me redo:

(1/2)∫(2s² - s + s/(s+1)²)ds = (1/2)[2s³/3 - s²/2 + ∫s/(s+1)² ds]

= s³/3 - s²/4 + (1/2)∫s/(s+1)² ds

∫s/(s+1)² ds: let u = s+1, s = u-1, ds = du ∫(u-1)/u² du = ∫(1/u - 1/u²)du = ln|u| + 1/u = ln(s+1) + 1/(s+1)

So F(s) = s³/3 - s²/4 + (1/2)ln(s+1) + 1/(2(s+1)) ✓

Let me recompute F(3) and F(2):

F(3) = 27/3 - 9/4 + (1/2)ln(4) + 1/(2·4) = 9 - 9/4 + (1/2)·2ln(2) + 1/8 = 9 - 2.25 + ln(2) + 0.125 = 6.875 + ln(2)

F(2) = 8/3 - 4/4 + (1/2)ln(3) + 1/(2·3) = 8/3 - 1 + (1/2)ln(3) + 1/6 = 8/3 - 5/6 + (1/2)ln(3) = 16/6 - 5/6 + (1/2)ln(3) = 11/6 + (1/2)ln(3)

F(3) - F(2) = 6.875 + ln(2) - 11/6 - (1/2)ln(3)

6.875 = 55/8, 11/6

55/8 - 11/6 = (165 - 44)/24 = 121/24

So F(3) - F(2) = 121/24 + ln(2) - (1/2)ln(3)

This doesn't match any answer choice either.

Let me reconsider the problem. Maybe the integral limits are not 4 and 27 in t, but rather the integral is with respect to something else, or maybe I need to reconsider.

Actually, wait. Let me re-read the problem statement more carefully.

"Evaluate Z 27 4 1 2 g(t)dt."

Hmm, maybe it's ∫[4,27] (1/2)g(t)dt, and the limits 4 and 27 correspond to s = 2 and s = 3?

If s = 2: t = 2³/(1+2) = 8/3 If s = 3: t = 3³/(1+3) = 27/4

Neither is 4 or 27. But what if I misread and the limits are 8/3 and 27/4? No, the problem clearly says 4 and 27.

Let me try another interpretation. What if the integral is ∫[4,27] (1/2)g(t)dt and the limits correspond to s values where t = 4 and t = 27?

For t = 4: s³/(1+s) = 4 → s³ - 4s - 4 = 0 For t = 27: s³/(1+s) = 27 → s³ - 27s - 27 = 0

Let me try to factor s³ - 4s - 4: Possible rational roots: ±1, ±2, ±4 s = 2: 8 - 8 - 4 = -4 ≠ 0 s = -2: -8 + 8 - 4 = -4 ≠ 0 No rational roots.

s³ - 27s - 27: Possible rational roots: ±1, ±3, ±9, ±27 s = 3: 27 - 81 - 27 = -81 ≠ 0 s = -3: -27 + 81 - 27 = 27 ≠ 0 No rational roots.

So the s values aren't nice. Let me reconsider the problem.

Hmm, maybe I should reconsider whether the integral is actually ∫[4,27] (1/2)g(t)dt where the answer is expressed in terms of the s values, and maybe the answer choices correspond to a different computation.

Actually, let me reconsider. Maybe the problem is asking for the integral and the answer is one of ①-⑤. Let me try to see if the answer could be ④ 79/6 + ln 2.

79/6 ≈ 13.167, ln 2 ≈ 0.693, total ≈ 13.86.

My numerical estimate was ≈ 49.03, which is way off. So I must be making an error.

Let me reconsider. Maybe I have the wrong f(x). Let me re-read: "f(x) = 1 2 x2 −x+ ln(1 +x)"

Could this be f(x) = 1/(2x² - x + ln(1+x))? No, that doesn't make sense.

Or f(x) = (1/2)x² - x + ln(1+x)? That's what I assumed.

Or maybe f(x) = 1/(2x²) - x + ln(1+x)?

Let me try f(x) = 1/(2x²) - x + ln(1+x): f'(x) = -1/x³ - 1 + 1/(1+x)

This seems messy and unlikely to give a clean answer.

Let me go back to f(x) = (1/2)x² - x + ln(1+x) and reconsider.

Actually, wait. Let me reconsider the integral. Maybe it's not ∫(1/2)g(t)dt but rather the integral has a different form. Let me re-read:

"Evaluate Z 27 4 1 2 g(t)dt."

In Korean math notation, this could be ∫[4,27] (1/2)g(t)dt. But maybe it's ∫[4,27] (1/2)·g(t)·dt, which is what I computed.

Hmm, let me try a completely different approach. Let me reconsider whether the distance might be different.

Actually, wait. Let me reconsider. The "foot of the perpendicular from (s, f(s)) to the y-axis" — is this (0, f(s))? The perpendicular to the y-axis (which is the line x=0) from point (s, f(s)) is the horizontal line, so the foot is indeed (0, f(s)). ✓

And the tangent meets the y-axis at (0, f(s) - s·f'(s)). ✓

Distance = |s·f'(s)| = s³/(1+s). ✓

Hmm, let me try yet another interpretation. What if the integral is with respect to s, not t? Like ∫[4,27] (1/2)g(t)dt where g(t) = s and t = s³/(1+s)?

Actually, I already did that. The integral becomes (1/2)∫s·dt = (1/2)∫s·(dt/ds)ds.

Let me try to see if the answer ④ 79/6 + ln 2 could come from s = 2 to s = 3 with some different integrand.

If the integral were ∫[2,3] (1/2)s·dt where t = s³/(1+s): = (1/2)∫[2,3] s · s²(3+2s)/(1+s)² ds = (1/2)∫[2,3] s³(3+2s)/(1+s)² ds

I computed F(3) - F(2) = 121/24 + ln(2) - (1/2)ln(3) ≈ 5.04 + 0.693 - 0.549 = 5.18

That doesn't match 13.86 either.

Let me try ∫[2,3] s³(3+2s)/(1+s)² ds (without the 1/2): = 2·(F(3) - F(2)) = 2·(121/24 + ln(2) - (1/2)ln(3)) = 121/12 + 2ln(2) - ln(3) ≈ 10.08 + 1.386 - 1.099 = 10.37

Still doesn't match.

Hmm, let me try a completely different approach. Maybe I should reconsider the problem from scratch.

Actually, let me reconsider. Maybe the integral is ∫[4,27] (1/2)g(t)dt and the limits 4 and 27 are values of s, not t. But that doesn't make sense because g is a function of t.

Or maybe the limits are values of g(t), i.e., s goes from 4 to 27? That would be a huge range.

Let me try: if s goes from 4 to 27, then t goes from 4³/5 = 64/5 = 12.8 to 27³/28 = 19683/28 ≈ 703.

The integral (1/2)∫[4,27] s · s²(3+2s)/(1+s)² ds = (1/2)[F(27) - F(4)]

F(27) = 27³/3 - 27²/4 + (1/2)ln(28) + 1/56 = 6561/3 - 729/4 + (1/2)ln(28) + 1/56 = 2187 - 182.25 + (1/2)ln(28) + 0.0179 = 2004.77 + (1/2)ln(28)

F(4) = 64/3 - 4 + (1/2)ln(5) + 1/10 = 21.333 - 4 + (1/2)ln(5) + 0.1 = 17.433 + (1/2)ln(5)

F(27) - F(4) ≈ 2004.77 - 17.433 + (1/2)(ln(28) - ln(5)) = 1987.34 + (1/2)ln(5.6) ≈ 1987.34 + 0.5·1.723 = 1988.2

(1/2)·1988.2 ≈ 994. Way too big.

OK, I think I need to reconsider the problem more carefully. Let me re-read the original problem statement.

"Consider the function f(x) = 1/2 x² - x + ln(1+x), and let t be a positive real number. For s > 0, let (s, f(s)) be a point on the curve y = f(x). Let g(t) denote the value of s such that the distance between the foot of the perpendicular from (s, f(s)) to the y-axis and the point where the tangent to the curve at (s, f(s)) meets the y-axis is equal to t. Evaluate ∫[4,27] (1/2)g(t)dt."

OK so my setup seems right. Let me reconsider the computation.

Actually, wait. Let me reconsider. Maybe I need to check: is f'(s) = s²/(1+s) correct?

f(x) = (1/2)x² - x + ln(1+x) f'(x) = x - 1 + 1/(1+x)

Let me verify: d/dx[(1/2)x²] = x, d/dx[-x] = -1, d/dx[ln(1+x)] = 1/(1+x). ✓

f'(x) = x - 1 + 1/(1+x) = [x(1+x) - (1+x) + 1]/(1+x) = [x + x² - 1 - x + 1]/(1+x) = x²/(1+x). ✓

So the distance is s·f'(s) = s³/(1+s). ✓

Hmm, let me try to think about this differently. Maybe the answer choices give a hint. Let me see if any of them could be F(s₂) - F(s₁) for nice s values.

④ 79/6 + ln 2: F(s₂) - F(s₁) = 79/6 + ln 2

If s₁ = 2, s₂ = 3: F(3) - F(2) = 121/24 + ln(2) - (1/2)ln(3) ≈ 5.04 + 0.693 - 0.549 = 5.18

79/6 ≈ 13.17. Not matching.

Let me try s₁ = 1, s₂ = 2: F(2) = 8/3 - 1 + (1/2)ln(3) + 1/6 = 11/6 + (1/2)ln(3) F(1) = 1/3 - 1/4 + (1/2)ln(2) + 1/4 = 1/3 + (1/2)ln(2)

F(2) - F(1) = 11/6 - 1/3 + (1/2)ln(3) - (1/2)ln(2) = 11/6 - 2/6 + (1/2)ln(3/2) = 9/6 + (1/2)ln(3/2) = 3/2 + (1/2)ln(3/2) ≈ 1.5 + 0.202 = 1.702

Not matching.

Let me try s₁ = 1, s₂ = 3: F(3) - F(1) = 9 - 9/4 + ln(2) + 1/8 - 1/3 - (1/2)ln(2) = 9 - 2.25 + 0.125 - 0.333 + ln(2) - (1/2)ln(2) = 6.542 + (1/2)ln(2) = 6.542 + 0.347 = 6.889

Not matching.

Hmm, let me try to see if the answer could be ④ with a different setup.

79/6 + ln 2 ≈ 13.167 + 0.693 = 13.86

Let me try s₁ = 2, s₂ = 4: F(4) = 64/3 - 4 + (1/2)ln(5) + 1/10 = 64/3 - 4 + 1/10 + (1/2)ln(5) = 64/3 - 39/10 + (1/2)ln(5) = (640 - 117)/30 + (1/2)ln(5) = 523/30 + (1/2)ln(5)

F(2) = 11/6 + (1/2)ln(3)

F(4) - F(2) = 523/30 - 11/6 + (1/2)ln(5) - (1/2)ln(3) = 523/30 - 55/30 + (1/2)ln(5/3) = 468/30 + (1/2)ln(5/3) = 78/5 + (1/2)ln(5/3) ≈ 15.6 + 0.255 = 15.855

Not matching.

Let me try s₁ = 2, s₂ = 3 with the full integral (without 1/2): 2·(F(3) - F(2)) = 121/12 + 2ln(2) - ln(3) ≈ 10.08 + 1.386 - 1.099 = 10.37

Not matching.

Hmm, let me try a completely different approach. Maybe I should reconsider the problem.

Actually, let me reconsider. Maybe the integral is not ∫[4,27] (1/2)g(t)dt but rather the problem has a different structure. Let me re-read the Korean.

"Z 27 4 1 2 g(t)dt" — in Korean math notation, this is ∫[4,27] (1/2)g(t)dt.

But wait, maybe it's ∫[4,27] (1/2)·g(t)·dt, and the answer is supposed to be computed using the substitution, and the limits 4 and 27 correspond to s values that I need to find.

Let me try to solve s³/(1+s) = 4 more carefully.

s³ - 4s - 4 = 0

Let me try s = 2: 8 - 8 - 4 = -4 s = 2.5: 15.625 - 10 - 4 = 1.625 s = 2.3: 12.167 - 9.2 - 4 = -1.033 s = 2.4: 13.824 - 9.6 - 4 = 0.224 s = 2.38: 13.485 - 9.52 - 4 = -0.035 s = 2.382: 13.52 - 9.528 - 4 = -0.008 s = 2.383: 13.54 - 9.532 - 4 = 0.008

So s₁ ≈ 2.3825

s³ - 27s - 27 = 0: s = 5.6: 175.616 - 151.2 - 27 = -2.584 s = 5.7: 185.193 - 153.9 - 27 = 4.293 s = 5.65: 180.36 - 152.55 - 27 = 0.81 s = 5.63: 178.45 - 152.01 - 27 = -0.56 s = 5.64: 179.4 - 152.28 - 27 = 0.12 s = 5.639: 179.3 - 152.25 - 27 = 0.05

So s₂ ≈ 5.639

Now F(s₂) - F(s₁): F(5.639) = 5.639³/3 - 5.639²/4 + (1/2)ln(6.639) + 1/(2·6.639) = 179.3/3 - 31.8/4 + 0.5·1.893 + 0.0753 = 59.77 - 7.95 + 0.947 + 0.0753 = 52.84

F(2.3825) = 2.3825³/3 - 2.3825²/4 + (1/2)ln(3.3825) + 1/(2·3.3825) = 13.53/3 - 5.676/4 + 0.5·1.218 + 0.1478 = 4.51 - 1.419 + 0.609 + 0.1478 = 3.848

F(s₂) - F(s₁) ≈ 52.84 - 3.848 = 48.99

(1/2)·48.99 ≈ 24.5

Hmm, none of the answer choices are close to 24.5.

Wait, I think I may be making an error. Let me reconsider.

Actually, let me reconsider the problem. Maybe the integral is ∫[4,27] (1/2)g(t)dt, and I should use the substitution differently.

We have t = s³/(1+s), so g(t) = s.

∫[4,27] (1/2)g(t)dt = ∫[4,27] (1/2)s dt

With dt = s²(3+2s)/(1+s)² ds:

= (1/2) ∫[s₁,s₂] s · s²(3+2s)/(1+s)² ds = (1/2) ∫[s₁,s₂] s³(3+2s)/(1+s)² ds

I computed the antiderivative as F(s) = s³/3 - s²/4 + (1/2)ln(s+1) + 1/(2(s+1)).

Let me verify by differentiating: F'(s) = s² - s/2 + 1/(2(s+1)) - 1/(2(s+1)²)

Let me check if this equals (1/2)·s³(3+2s)/(1+s)²:

(1/2)·s³(3+2s)/(1+s)² = s³(3+2s)/(2(1+s)²)

F'(s) = s² - s/2 + 1/(2(s+1)) - 1/(2(s+1)²) = [2s²(1+s)² - s(1+s)² + (1+s) - 1] / [2(1+s)²] = [2s²(1+2s+s²) - s(1+2s+s²) + 1+s - 1] / [2(1+s)²] = [2s² + 4s³ + 2s⁴ - s - 2s² - s³ + s] / [2(1+s)²] = [2s⁴ + 3s³] / [2(1+s)²] = s³(2s+3) / [2(1+s)²]

Yes! This matches. ✓

So the antiderivative is correct. The issue must be with the limits.

Let me reconsider. Maybe the problem is not ∫[4,27] (1/2)g(t)dt but something else. Let me re-read the problem statement once more.

"Evaluate Z 27 4 1 2 g(t)dt."

Hmm, what if this is actually ∫[4,27] (1/2)g(t)dt where the limits are values of s, not t? That is, the integral is with respect to t, but the limits 4 and 27 are values of s = g(t)?

No, that doesn't make sense. The limits of an integral with respect to t should be values of t.

Wait, actually, maybe I should reconsider. What if the problem is asking for the integral from t = 4 to t = 27, and the answer involves ln 2 or ln 3, and the answer choices are correct?

Let me try to compute more carefully.

F(s) = s³/3 - s²/4 + (1/2)ln(s+1) + 1/(2(s+1))

For t = 4: s³/(1+s) = 4, so s³ = 4(1+s) = 4 + 4s For t = 27: s³/(1+s) = 27, so s³ = 27(1+s) = 27 + 27s

Let me use the relationship s³ = t(1+s) to simplify F(s):

F(s) = t(1+s)/3 - s²/4 + (1/2)ln(s+1) + 1/(2(s+1))

This still depends on s, not just t. So the integral F(s₂) - F(s₁) depends on the specific s values.

Let me try to compute F(s₂) - F(s₁) using the relationship s³ = t(1+s).

F(s) = s³/3 - s²/4 + (1/2)ln(s+1) + 1/(2(s+1)) = t(1+s)/3 - s²/4 + (1/2)ln(s+1) + 1/(2(s+1))

For s₁ (t=4): F(s₁) = 4(1+s₁)/3 - s₁²/4 + (1/2)ln(1+s₁) + 1/(2(1+s₁)) For s₂ (t=27): F(s₂) = 27(1+s₂)/3 - s₂²/4 + (1/2)ln(1+s₂) + 1/(2(1+s₂)) = 9(1+s₂) - s₂²/4 + (1/2)ln(

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