hw asst 21

course mth 272

??????K?+????}?assignment #021

021.

Applied Calculus II

07-23-2008

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20:44:17

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

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RESPONSE -->

first you apply the definition of the improper integral

integral from -infinity to infinity of x^2 e^(-x^3) =

integral from -infinity to c of x^2 e^(-x^3) + integral from c to infinity of x^2 e^(-x^3)

antiderivative =

x^3 / 3 * (e^-x^3) (-3x)

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

apply fundamental theorem with -infinity to c and c to infinity

answer = the integral will diverge bc -infinity to c will diverge and it only takes one of the integrals on the right to make the integral on the left diverge.

confidence assessment: 2

** 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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20:44:48

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

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RESPONSE -->

No the integral does not converge. This was explained in previous answer.

confidence assessment:

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21:07:11

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 -->

= integral from 0 to 20 of 75,000e^-.08t dt

= [(75,000 / -.08)e^-0.08t] at 0 and 20 or at 0 and infinity

apply fundamental theorem

for 20 years

for 20 =371477.4318

for 0 = 75000

=$296,477

for forever

apply fundamental theorem

for infinity = b = 0

for 0 =75000

0 - 75000 = -$75000

confidence assessment: 2

For reference, not that $75,000 per year for 20 years totals $1.5 million. At 8% you would need less than $1.5 million today to equal the income from the farm.

The correct integral for the present value is INT (75,000 e ^ (-.08 * t ) dt, t from 0 to 20).

An antiderivative is 75,000 /( -0.08 ) * e^ (-0.08t) . This agrees with your antiderivative.

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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21:07:29

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

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RESPONSE -->

already answered

confidence assessment:

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21:08:07

What integrals did you evaluate to get your results?

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RESPONSE -->

already answered

0 to 20

and 0 to infinity which is 0 to b

confidence assessment: 2

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Your work looks good. A few errors in detail, but overall very solid. See my notes.