hw asst 19

course mth 272

ԗy򴁈Ƌassignment #019

019.

Applied Calculus II

07-20-2008

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16:06:35

Query problem 6.4.16 use table to integrate x^2 ( ln(x^3) )^2

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RESPONSE -->

i dont see any formulas that work for this problem.

Question: are you able to combine formulas. - for the x^2 part and the (ln(x^3) )^2 part -bc I do see formulas for those.

use formula # 1 for the x^2

= x^3 /3

use formula #42 for the ln(x^3) )^2 part

= u[2- 2 ln u + (ln u)^2] + C

= x^3[2 - 2 ln(x^3) + ln(x^3)^2] + C

add the (x^3 /3) part

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

confidence assessment: 1

The integral of f g is not equal to integral of f * integral of g (since (fg)' = f ' g + g ' f you wouldn't expect it to work out for integrals either).

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.

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16:06:55

What is your result?

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RESPONSE -->

already answered

confidence assessment: 3

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16:07:24

What formula did you use from the table and how did you use it?

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RESPONSE -->

i used formula 1 and formula 42

confidence assessment:

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16:23:00

Query problem 6.4.46 use table to integrate x ^ 4 ln(x) then check by integration by parts

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RESPONSE -->

want to use formula 41

#41 = u^n ln u = u^n +1 / (n+1)^2 [-1 + (n+1) ln u] + C

x ^ 4 ln(x)

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

checking by integration by parts...

dv = x^4 dx v = integral of dv = integral of x^4 dx =x^5/5

u = ln x du = 1/x dx

integral of x ^ 4 ln(x)

= x^5 /5 ln x - integral of (x^5/5)(1/x) dx

=x^5 /5 ln x - 1/5 integral of x^4 dx

=x^5/5 ln x - (x^5/15) + C

confidence assessment: 2

The 15 in your last step should be 25; likely typo because otherwise everything is done correctly.

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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16:23:10

What is your integral?

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RESPONSE -->

already answered

confidence assessment:

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16:23:22

What formula did you use from the table and how did you use it?

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RESPONSE -->

formual 41

confidence assessment:

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16:25:42

Explain how you used integration by parts to obtain your result.

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RESPONSE -->

already shown in work.

i let dv =x^4 dx =x^5 / 5

i let u = ln x du = 1/x dx

then solved using formula...

integral of ( u dv = uv - integral of v du)

confidence assessment: 2

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16:43:26

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

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RESPONSE -->

looking at the tables, I can't find an accepable formula

=integral of 'sqrt( 375.6 t^2 - 715.86)

= (375.6 t^2 - 715.86)^2

= (375.6 t^2 - 715.86)^3 / 3

on [8,16]

use fundamental theorem

=(375.6 (16)^2 - 715.86)^3 / 3

= really big number. probably not right then

=(375.6 (8)^2 - 715.86)^3 / 3

= really big number. prob not right then

add them together

confidence assessment: 1

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).

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

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

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. *&*&

note on approximate comparison: 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.

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16:47:44

What is the average net profit over the given time interval?

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RESPONSE -->

plugging into original formula with out integration might work.

plugging in 16 - 8

t = 16 = 308.93

t = 8 = 152.71

308.93 - 152.71 = 156.21

confidence assessment: 1

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16:47:59

Explain how you obtained your result.

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RESPONSE -->

shown in previous work

confidence assessment:

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This looks good. See my notes. Let me know if you have any questions. &#