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course mth 272

иl·—~”ˆÓèÇ£ÿãN‚ëýëòyassignment #016

016.

Applied Calculus II

07-17-2008

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12:59:52

Query problem 6.2.50 (was 6.2.48) solid of revolution y = x e^x x = 0 to 1 about x axis

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RESPONSE -->

dv = e^x dx v =integral of dv = integral of e^x dx = e^x

u = x du=dx

x e^x = x e^x - integral of e^x dx

= x e^x - e^x + C

use fundamental theorem at x= 0 and x=1

(1) e^(1) - e^(1) + C =0

(0) e^(0) - e^(0) + C = -1

= -1

confidence assessment: 2

The volume over an interval `dx will be approximately equal to pi ( x e^x)^2 `dx, so we will be integrating pi x^2 e^(2x) with respect to x from x = 0 to x = 1.

Integrating just x^2 e^(2x) we let u = x^2 and dv = e^(2x) dx so that du = 2x dx and v = 1/2 e^(2x).

This gives us u v - int(v du) = 2x (1/2 e^(2x)) - int(1/2 * 2x e^(2x) ). The remaining integral is calculated by a similar method and we get a result that simplifies to

e^(2x) ( 2 x^2 - 2x + 1) / 4.

Our antiderivative is therefore pi e^(2x) (2 x^2 - 2x + 1) / 4. This antiderivative changes between x = 0 and x = 1 by pi ( e^2 / 4 - 1/4) = 5.02, approx..

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13:00:36

What expression do you integrate to obtain the desired result?

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RESPONSE -->

x e^x - integral of e^x dx

confidence assessment: 3

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13:00:46

What is your result?

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RESPONSE -->

-1

confidence assessment: 2

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13:00:59

For your first step, what was u and what was dv, what were du and v, and what integral did you obtain?

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RESPONSE -->

already answered

confidence assessment: 2

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13:02:09

Answer the same questions for your second step.

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RESPONSE -->

? so is this a using integration by parts repeatedly problem then?

confidence assessment: 0

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13:05:34

Problem 6.2.58 (was 6.2.56) revenue function 410.5 t^2 e^(-t/30) + 25000

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RESPONSE -->

ok

confidence assessment:

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13:20:37

What expression did you evaluate to obtain the average daily receipts during the first quarter, and what was your result?

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RESPONSE -->

use

integral from x= 0 to x=90 of 410.5t^2 e^-t/30 + 25,000

dv= e^-t/30 v =integral of dv = integral of e^-t/30 = -1/30e^-t/30

u = t^2 du= 2dt

integral of t^2e^-t/30 dt = -1/30e^-t/30 + -1/30 integral of e^-t/30 dt

=-1/30e^-t/30 - 30e^-t/30

= e^t/30

pv = 410.5 integral at x=0 and 90 of (e^-t/30)

=20.437

=410.5

20.437 -410.5 = -390.06

confidence assessment:

If we integrate the revenue function from t = 0 to t = 90 we will have the first-quarter revenue. If we then divide by 90 we get the average daily revenue.

To get an antiderivative of t^2 e^(-t/30) we first substitute u = t^2 and dv = e^(-t/30), obtaining du = 2t dt and v = -30 e^(-t/30). We proceed through the rest of the steps, which are very similar to steps used in preceding problems, to get antiderivative

- 30•e^(- t/30)•(t^2 + 60•t + 1800).

Our antiderivative of 410.5 t^2 e^(-t/30) + 25000 is therefore 410.5 (- 30•e^(- t/30)•(t^2 + 60•t + 1800) ) + 25000 t.

The change in this antiderivative function between t = 0 and t = 90 is found by substitution to be about 15,000,000, representing $15,000,000 in 90 days for an average daily revenue of $15,000,000 / 90 = $167,000, approx..

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13:23:05

What expression did you evaluate to obtain the year's total daily receipts and was your result?

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RESPONSE -->

you use the same expression for teh fourth quarter and for the total daily receipts as you did in the first quarter.

your x values change however

x= 274 and x=365

for total year

x=0 and x=365

confidence assessment: 1

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13:35:32

Problem 6.2.64 (was 6.2.62) c = 5000 + 25 t e^(-t/10), r=6%, t1=10 yr, find present value

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RESPONSE -->

integral from x=0 and x=10 of 5025 e^.06(-t/10)

You don't add the 25 to 5000. Order of operations. The 25 is multiplied the t e^(-t/10), not added to 5000.

=(5025 / .06) e^.06(-t/10) from 0 to 10

use fundamental theorem.

= 78872.78 - 83750

= -4877.22

don't think this is right because i dont think I should have a negative answer.

confidence assessment: 2

The income during a time interval `dt is ( 5000 + 25 t e^(-t/10) ) `dt. To get this amount after t years we would have to invest ( 5000 + 25 t e^(-t/10) ) `dt * e^(-.06 t).

Integrating this expression from t = 0 to t = 10 we obtain

int(( 5000 + 25 t e^(-t/10) ) * e^(-.06 t) dt , t from 0 to 10).

Our result is $38,063.

Note that the entire income stream gives us int(( 5000 + 25 t e^(-t/10) ) dt , t from 0 to 10) = $50,660 over the 10-year period.

The meaning of our solution is that an investment of $38,063 for the 10-year period at 6% continuously compounded interest would yield $50,660 at the end of the period. So $38,063 is the present value of the income stream.

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13:35:51

What expression do you integrate to obtain the present value of the income function and what is your result?

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RESPONSE -->

already in previous answer

confidence assessment: 2

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13:37:48

Explain the meaning of the expression you integrated--why does this function give the present value?

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RESPONSE -->

this present value represents the amount of income you have now that is going to grow over the next 10 years.

confidence assessment: 2

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