Improper Integrals and Separable ODEs, calculus homework help

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Name: Recitation Time: Math 252 Improper Integrals and Separable ODEs Activity 7 This activity is worth 10 points of course credit. See tentative calendar for due dates. Late activities are accepted at the discretion of your recitation instructor and a penalty may be imposed. (1) Let’s test your intuition. Is the following statement true or false? Explain (with complete and grammatically correct sentences) your reasoning. If an unbounded solid object has finite volume then any cross-section of the object must have finite area. Note: Credit will be given for any coherent answer (regardless of whether it’s correct). (2) Gabriel’s Horn! Consider the unbounded region R under the curve y = the x-axis) on the interval [1, ∞). 1 (and above x (a) Show via an improper integral that the area of R is infinite. (b) Show via an improper integral that the volume of the solid obtained by revolving this region about the x-axis is finite and report the volume of this unbounded solid object. V = Z ∞ (3) For what value(s) of p > 0 is integral (as a function of p). 1 1 dx finite? For those value(s), determine the xp Z ∞ For the value(s) of p above, 1 1 dx = xp Z 1 (4) For what value(s) of p > 0 is integral (as a function of p). 0 1 dx finite? For those value(s), determine the xp Z 1 For the value(s) of p above, 0 1 dx = xp (5) Solve each of the following separable ODEs: (a) y 0 + xy = 2x. y= (b) xy 0 = tan (y). y= (c) y 0 = 1 − y2 where y is a function of t. 2 y= (6) A tank initially contains 30 lbs of salt dissolved in 1, 000 gallons of water at time t = 0. Brine that contains 0.2 lbs of salt per gallon is pumped into the tank at 5 gallons per minute and the well-mixed solution is drained at the same rate. (a) Determine a function y(t) for the amount of salt in the tank (in lbs) at time t ≥ 0. Hint: y 0 (t) = rate (lbs/min) of salt coming - rate (lbs/min) of salt going out. y(t) = (b) Determine the equilibrium solution, that is, lim y(t). t→∞ lim y(t) = t→∞
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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...


Anonymous
I was having a hard time with this subject, and this was a great help.

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