Assignment 21

course Mth 272

?_??????}?z}assignment #021

021.

Applied Calculus II

04-02-2009

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00:12:03

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

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RESPONSE -->

from -inf to inf finding the integral first

u=-x^3

du= -3X^2dx

-1/3~ e^-x^3dx = -1/3e^-x^3 from -inf to inf

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

0+ inf = Infinite, it diverges since it doesn't have a definite answer.

confidence assessment: 3

Good. Infinity isn't a number so you can't substitute it as a limit, so see below for an example of how to handle this part of the solution more rigorously.

** The integral as stated here 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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00:24:13

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

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answered

confidence assessment: 3

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00:39:15

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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RESPONSE -->

a.) pv= ~from 0 to 20. 75000e^-.08t dt

u=-.08t

du= -.08dt

-75000/.08 ~ e^-.08tdt

-75000/.08 (e^-.08t) from 0 to 20

-75000/.08 [e^-.08(20) - e^-.08(0)] = -75000/.08[.2019-1]

= 748222.0144

b.) from 0 to inf. -75000/.08 e^-.08t

-75000/.08 [e^-.08(inf) - e^^-.08(0)] = -75000/.08 [0-1]

= $937500

the results show that after 20 years, there is not much of a significant increase in the farms profit.

confidence assessment: 3

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00:39:23

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

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answered

confidence assessment: 3

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00:39:30

What integrals did you evaluate to get your results?

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RESPONSE -->

answered

confidence assessment: 3

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00:39:57

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

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RESPONSE -->

I pretty much understood all of this, i don't have any questions on this one. Thanks.

confidence assessment: 3

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&#Good work. See my notes and let me know if you have questions. &#