Assignment 21

course MTH 272

......!!!!!!!!...................................Applied Calculus II

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Asst # 21

07-04-2006

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14:28:57

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**** Query problem 6.6.14 integral from -infinity to infinity of x^2 e^(-x^3)

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14:34:12

u= -x^3

du= -3x^2

= -1/3 int (e^(-x^3)

= 1/3 e^(-x^3)

(1/3e^(-x)) from [-inf, 0] + (1/3e^(-x)) from [0, inf]

(1/3e^0 - 1/3e^inf) + (1/3e^(-inf) - 1/3e^0)

=(1/3 - 0) + (0 - 1/3)

= 1/3 - 1/3

However, I think the integral diverges since it diverges on the negative infinity side, and if it diverges, does that mean you're unable to find the limit? Also, how can you tell if it diverges without looking at a graph of the function?

In this case the graph would make it obvious, but the graph doesn't always help. For example int(1/x^1.01, x from 1 to infinity) converges, but if it was 1 / x^.99 in would diverge, and the graphs don't look much different.

You need to use a limiting process to determine whether a given integral is convergent or divergent. The solution given below should help illustrate the process:

The integral in the present problem diverges.

You need to take the limit as t -> infinity of INT(x^2 e^(-x^3), x from -t to t ).

Using the obvious substitution we see that the result is the same as the limiting value as t -> infinity of INT( 1/3 e^(-u), u from -t to t ). Using -1/3 e^(-u) as antiderivative we get -1/3 e^(-t)) - (-1/3 e^(-(-t))); the second term is 1/3 e^t, which approaches infinity as t -> infinity. The first term approaches zero, but that doesn't help. The integral approaches infinity.

Note that the integral from 0 to infinity converges: We take the limit as t -> infinity of INT(x^2 e^(-x^3), x from 0 to t ), which using the same steps as before gives us the limit as t -> infinity of -1/3 e^(-t) - (-1/3) e^0. The first term approaches zero, the second is just 1/3. So the limiting value is 1/3. **

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14:34:12

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**** Does the integral converge, and if so what is its value? Explain why the integral does or does not converge.

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14:35:05

I think the integral diverges because y approaches infinity and x approaches negative infinity.

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14:35:06

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**** Query problem 6.6.40 (was 6.6.38) farm profit of $75K per year, 8% continuously compounded, find present value of the farm for 20 years, and forever.

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15:03:08

75,000 int(e^(-.08t))

= 75,000 (-25/2) e^(-.08t)

= -937,500e^(-.08t)

a) over 20 years: [0,20]

=(-937,500e^(-.08*20)) - (-937,500e^0)

=-189,277.9856 + 937,500

= 748,222.01

b) forever: [0, infinity]

= (-937,500e^(-.08*infinity)) - (-937,500e^0)

= 0 + 937,500

=937,500

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15:03:08

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**** What is the present value of the farm for 20 years, and what is its present value forever?

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15:03:14

75,000 int(e^(-.08t))

= 75,000 (-25/2) e^(-.08t)

= -937,500e^(-.08t)

a) over 20 years: [0,20]

=(-937,500e^(-.08*20)) - (-937,500e^0)

=-189,277.9856 + 937,500

= 748,222.01

b) forever: [0, infinity]

= (-937,500e^(-.08*infinity)) - (-937,500e^0)

= 0 + 937,500

=937,500

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15:03:14

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**** What integrals did you evaluate to get your results?

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15:03:45

I evaluated:

75,000 int(e^(-.08t))

= 75,000 (-25/2) e^(-.08t)

= -937,500e^(-.08t)

I used the same integral for both parts

Your solution agrees with mine. Good work.

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15:03:45

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**** Query Add comments on any surprises or insights you experienced as a result of this assignment.

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15:05:15

I found myself really struggling to understand this section on improper integrals. I've looked at the class notes on the CD, but I think I'm going to have to go back over those, and hopefully things will start to sink in. There isn't one specific thing I don't understand- I think it's just that in general, I don't understand improper integrals, the need to solve them a special way, etc.

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See my notes on that one improper integral, and keep in mind that you generally need to make at least one of the 'ends' of the interval of integration into a variable and let it approach its limit. Let me know if you have questions.