Assignment 19

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course MTH272

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.

019.

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Question: `qQuery problem 6.4.16 use table to integrate x^2 ( ln(x^3) )^2

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Your solution:

x^2 ( ln(x^3) )^2

using #42 [(int.) u (ln u)^2 du=u(2-2(ln u) + (ln u)^2)+C]

u=x^3

du=3x^2

X^2 = 1/3

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du = 3 x^2 dx, not just 3 x^2.

So x^2 dx = 1/3 du.

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x^3[(2-2 ln(x^3)+(ln(x^3)^2)]/3

[(2 x^3)-(2 x^3)( ln(x^3)+(ln(x^3)^2)]/3

confidence rating #$&*:1

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Given Solution:

`a Let u = x^3 so du = 3x^2 dx. The x^2 dx in the integral is just 1/3 du.

You therefore have the integral of 1/3 ( ln(u) )^2 du.

The table should have something for ( ln(u) ) ^ n.

In any case the integral of ln(u)^2 with respect to u is u ln(u)^2 - 2 u ln(u) + 2 u.

With the substitution u = x^3 you would be integrating (ln u)^2 * du/3, which would give you

u [ 2 - 2 ln u + (ln u)^2 ] / 3, which translates to

x^3 ( 2 - 2 ln(x^3) + (ln(x^3) ) ^ 2 ) / 3.

DER: int( (ln(u)^2) = u•LN(u)^2 - 2•u•LN(u) + 2•u. Then for increasing powers of n int( ln(u)^n) gives us:

u•LN(u)^3 - 3•u•LN(u)^2 + 6•u•LN(u) - 6•u then

u•LN(u)^4 - 4•u•LN(u)^3 + 12•u•LN(u)^2 - 24•u•LN(u) + 24•u and

u•LN(u)^5 - 5•u•LN(u)^4 + 20•u•LN(u)^3 - 60•u•LN(u)^2 + 120•u•LN(u) - 120•

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Self-critique (if necessary):I tried my best to get this solved but this is as far as I saw it working

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Self-critique Rating:2

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The integral becomes

int( ln(u)^2 du).

You will have a table entry for that case, probably in the form of the integral

int(ln(u)^n du).

Use that formula with n = 2.

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Question: `qQuery problem 6.3.52 (7th edition 6.4.46) use table to integrate x ^ 4 ln(x) then check by integration by parts

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Your solution:

#41from the tables in the book

u=ln(x)

du/dx = 1/x

v=x^(n+1)/(n + 1).

dv = x^n dx

[(u^(n+1))/(n+1)^2][-1+(n+1)(lnu) using #41 from tables

[ln(x) ^(5)/(5)^2][-1+(5)( ln(x))]

[ln(x) ^(5)/(25)][ (5 ln(x)-1/(5)]

[ln(x) ^(5)/(25)][ (4 ln(x)]+C

u v-(int. v du)

ln(x) x^5/5 - (int. (x^5/5)(1/x)

ln(x) x^5/5 - (int. (x^4/5)

(ln(x) x^5)/5 - (int. (x^4/5)

[(ln(x) x^5)/5] (x^5)/(25)

confidence rating #$&*:2

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Given Solution:

`a Integration by parts on x^n ln(x) works with the substitution

u = ln(x) and dv = x^n dx, so that

du/dx = 1/x and v = x^(n+1) / n, giving us

du = dx / x and v = x^(n+1) / (n + 1).

Thus our integral is

u v - int( v du) =

ln(x) * x^(n+1) / (n + 1) - int ( x^(n+1)/(n+1) * dx / x) =

ln(x) * x^(n+1)/( n + 1) - int(x^n dx) / (n+1) =

ln(x) * x^(n+1) / (n + 1) - x^(n+1) / (n+1)^2 =

x^(n+1) / (n+1) ( ln(x) - 1(n+1)).

This should be equivalent to the formula given in the text.

For n = 4 we get

x^(4 + 1) / (4 + 1) ( ln(x) - 1 / (4 + 1)) =

x^5 / 5 (ln(x) - 1/5). *&*&

(x^5/25)(4 ln x) + C

Using integration by parts:

(x^5/5) ln x - (x^5/25)

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Self-critique (if necessary):NA

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Self-critique Rating:

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Question: `qQuery problem 6.4.63 profit function P = `sqrt( 375.6 t^2 - 715.86) on [8,16].

( 8th edition problem is 6.2.61 and the function is sqrt(.000645 t^2 + .1673) on (2, 5) )

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Your solution:

P=sqrt(.000645 t^2 + .1673) on (2, 5) )

u = `sqrt(.00645)t^2 u = .08t rounded

@&

You can't ignore the .1673, which is also part of the expression of which the square root is taken. So this won't work.

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du or a=sqrt.1673 a=.41 rounded

using # 21 from the tables in the book

{sqrt(u^2 + or - a^2 du) = 1/2u (sqrt(u^2+ or - a^2)) + or - (a^2 ln|u) + (sqrt(u^2+ or - a^2))|

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.000645 t^2 + .1673 is

.000645 ( t^2 + .1673 / .000645)

= .000645 ( t^2 + 270),

very roughly.

So

sqrt(.000645 t^2 + .1673) =

sqrt(.000645) sqrt(t^2 + 270) =

.026 sqrt(t^2 + 270).

sqrt(t^2 + 270) is of the form

sqrt(t^2 + a^2)

if a = sqrt(270) = 16.5, roughly.

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}

½ {.08t (sqrt(.08t ^2+ .41^2)) + (.41^2 ln|.08t) + (sqrt(.08t ^2+.41^2))|}

2=.685

5=.700

Difference = .015 billions of dollars per year

confidence rating #$&*:3

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Given Solution:

`a To get the average net profit integrate the profit function over the given interval and divide by the length of the interval.

The integrand is sqrt(375.6 t^2 - 715.86), which is of the form `sqrt( u^2 +- a^2).

• By the table the integral is

1/2 u sqrt(u^2 +- a^2) -+ a^2 ln(u + sqrt(u^2 +- a^2)) + c.

u^2 = 375.67 t^2 so u = `sqrt(375.67) * t = 19.382 t approx.

Similarly a = `sqrt(715.86) = 26.755 approx..

Substituting our values of a and u we find that integral(sqrt(375.6 t^2 - 715.86), t from 8 to 16) will be about 1850. Dividing this by the length 8 of the interval gives us the average value, which is about 1850 / 8 = 230.

** THE FUNCTION IS CLOSE TO THE LINEAR FUNCTION 19.4 t. The 715.86 doesn't have much effect when t is 8 or greater so the function is fairly close to P = 19 t. This approximation is linear so its average value will occur at the midpoint t = 12 of the interval. At t = 12 we have P = 19 * 12 = 230, approx.

8TH EDITION INTEGRAL

The function given in the 8th edition leads to the integral

int( sqrt(.000645 t^2 + .1673) dt, t from 2 to 5), which is of the form `sqrt( u^2 +- a^2).

u^2 = .000645 t^2 so u = `sqrt(.000645) * t = .08 t approx.

Similarly a = `sqrt(.1673) = .41 approx..

• By the table the integral is

1/2 u sqrt(u^2 +- a^2) -+ a^2 ln(u + sqrt(u^2 +- a^2)) + c.

u^2 = 375.67 t^2 so u = `sqrt(375.67) * t = 19.382 t approx.

Similarly a = `sqrt(715.86) = 26.755 approx..

Substituting our values of a and u we find that

integral(sqrt(.000645 t^2 + .1673), t from 2 to 5) will be about 1850. Dividing this by the length 8 of the interval gives us the average value, which is about 1850 / 8 = 230.

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Self-critique (if necessary): I tried to look at how you were getting these numbers because that don’t match anything I got except the approximations.

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Self-critique Rating:

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

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Self-critique (if necessary):

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Self-critique rating:

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

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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Check my notes and let me know if you have additional questions.

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