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Solve the questions in the attachment, and provide explanation of the solution please.
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Explanation & Answer
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1. I suppose it is not true. It is similar, as I think, to the statement “If an integral of a function is finite,
then the function and the interval of integration must be bounded”, which is false as we know.
The value of a function is similar to the area of the cross-section, the integral (the area under the graph)
is similar to the volume of a solid.
2. Gabriel’s Horn.
a. The area of this unbounded figure is the limit of the area of the bounded figure (for 𝑥 between 1 and
𝐴) as 𝐴 tends to infinity. It is, in turn, equal to
𝐴
𝐴
2𝜋 ∫ 𝑓(𝑥)√1 +
2
(𝑓 ′ (𝑥)) 𝑑𝑥
1
1
It can be estimated from below:
1
1
√1 + 2
𝑥
𝑥
𝐴1
1
≥ 𝑥 and 2𝜋 ∫1
𝑥
1
1
= 2𝜋 ∫ √1 + 2 𝑑𝑥 .
𝑥
𝑥
1
𝑥2
√1 +
1
1
≥ 0, 1 + 𝑥 2 ≥ 1, √1 + 𝑥 2 ≥ 1. Therefore
𝐴1
1
𝑑𝑥
𝑥2
≥ 2𝜋 ∫1
𝑥
𝑑𝑥 = 2𝜋 ln 𝐴, which tends to infinity as 𝐴 → +∞.
Therefore the initial integral also tends to infinity as 𝐴 → ∞ and the area of Gabriel’s Horn is infinite.
b. The volume is the integral
∞
𝑉 = 𝜋∫
∞
(𝑓(𝑥))2
1
𝑇
1
1
1
𝑑𝑥 = 𝜋 ∫ 2 𝑑𝑥 = 𝜋 lim ∫ 2 𝑑𝑥 = 𝜋 lim (1 − ) = 𝝅.
𝑇→∞
𝑇→∞
𝑥
𝑥
𝑇
1
1
3. For 𝑝 ≠ 1 the integral is
∞
𝑇
𝑇
1
1
𝑇 1−𝑝 − 1
∫ 𝑝 𝑑𝑥 = lim ∫ 𝑝 𝑑𝑥 = lim ∫ 𝑥 −𝑝 𝑑𝑥 = lim
.
𝑇→∞
𝑇→∞
𝑇→∞ 1 − 𝑝
𝑥
𝑥
1
1
1
The limit is finite for 1 − 𝑝 < 0, 𝑝 > 1 (for those 𝑝 𝑇 1−𝑝 → 0, for 𝑝 < 1 𝑇 1−𝑝 → ∞).
The limit is
1
.
𝑝−1
𝑇1
For 𝑝 = 1 the limit is lim ∫1
𝑥
𝑇→∞
𝑑𝑥 = lim ln 𝑇 = ∞ (the integral diverges).
∞ 1
𝑑𝑥
𝑥𝑝
This way the final answer is ∫1
𝑇→∞
=
𝟏
𝒑−𝟏
for 𝒑 > 𝟏.
4. For 𝑝 ≠ 1 the integral is
1
1
1
1
1
1 − 𝑇 1−𝑝
−𝑝
∫ 𝑝 𝑑𝑥 = lim+ ∫ 𝑝 𝑑𝑥 = lim ∫ 𝑥 𝑑𝑥 = lim+
.
𝑇→0+
𝑇→0
𝑇→0
𝑥
𝑥
1−𝑝
0
𝑇
𝑇
The limit is finite fo...