21 open query

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course Mth 272

11/1 9:30 pm EST

021.

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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)

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Your solution:

U = x^3

Du = 3x^2 dx

1/3 [int.] e^-u du

1/3 [int.] –e^-u

1/3 [int.] –e^-x^3 = -e^-x^3 / 3

-1/3 [ e^- inf – e^inf ]

e^-inf = e^-1/0

-1/3 (0-inf)

-1/3 (-inf)

confidence rating #$&*: 3

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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.

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Self-critique (if necessary):

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self-critique rating #$&*: ‘OK”

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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.

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Your solution:

We have:

75000e^.08t

u = .08 t

du = .08 dx

e^ax/(ax) = e^ax/a

= 973,500e^0.8t / 0.08

If you used e^(0.8 t) rather than e^(.08 t), you will have obtained an erroneous result.

And in fact you should have used e^(-.08 t).

b = 20, a = 0

= 48,103, 463

confidence rating #$&*: 2

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Given Solution:

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

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.

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Self-critique (if necessary):

#### 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.

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).

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self-critique rating #$&*:

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

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Your solution:

Present value forever = 973,500

20 year present value = 48, 103,463 (like I mentioned, this is obviously wrong as it seems the value added upon the forever value through evaluation.)

confidence rating #$&*: 2

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

Forever $ 937,500.00

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Self-critique (if necessary):

#### Please see the previous correlating problem self-critique.

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self-critique rating #$&*: 2

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

That last problem gave me some worry, looking forward to curing my constraint there."

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

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