If your solution to a stated problem does not match the given solution, you should self-critique per instructions at  

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

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Question: `qQuery problem 6.5.22 (7th edition 6.6.14) integral from -infinity to infinity of x^2 e^(-x^3)

Your solution:

Confidence Assessment:

Given Solution:

`a 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.

Self-critique (if necessary):

Self-critique Rating:

Question: `qQuery problem 6.5.50 (7th edition 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.

Your solution:

Confidence Assessment:

Given Solution:

`a The correct antiderivative is [75,000 /-0.08 e^ -0.08t].

For 20 years you evaluate the change in this antiderivative between t = 0 and t = 20, and I believe you obtain $ 748,222.01

To get the present value forever you integrate from 0 to b and let b -> infinity.

The integral from 0 to b is 75,000 / (-.08) e^(-(.08 b)) - 75,000 / (-.08) e^(0.08 * 0) = 75,000 / -.08 * (e^(-.08 b) - e^0).

e^0 is 1 and as b -> infinity e^(-.08 b) -> 0. So the integral is

75,000 / -.08 ( 0 - 1) = 75,000 / .08 = 937,500.

STUDENT QUESTION

Curious how you got the antiderivative in the given solution, and the result after evaluation. I did something wrong here, as my main intention was to find the antiderivative of : 75000e^.08t, then evaluate; however, I think what I ended up doing was evaluating an antiderivative at the forever present value, and 20 years added on. Clarity to relieve my doubt is much needed. Thank you.

INSTRUCTOR RESPONSE

The present value of the income stream is integral(75 000 e^(-.08 t) dt).

e^-(.08 t) dt is integrated by letting u = -.08 t, so that du = 1/(-.08) dt and dt = -12.5 du.

We get the integral of -12.5 e^u du, which is -12.5 e^u. Translated back into t units this is -12.5 e^(-08 t).

Self-critique (if necessary):

Self-critique Rating:

Question: `qWhat is the present value of the farm for 20 years, and what is its present value forever?

Your solution:

Confidence Assessment:

Given Solution: `a The present value for 20 years is $ 748,222.01

Forever $ 937,500.00

Self-critique (if necessary):

Self-critique Rating:

Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.