Assignment 19

course Mth 272

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16:56:11

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

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Dont know which formula to use. I dont see one that is similar to this.

confidence assessment: 0

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16:56:14

What is your result?

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confidence assessment: 0

You do need to do a substitution to get this into the required general form:

** 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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16:56:17

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

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confidence assessment: 0

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16:56:24

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

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confidence assessment:

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16:56:37

What is your integral?

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

confidence assessment: 3

*&*& Note that 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). *&*&

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16:56:50

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

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Used formaula 41 int. u^n lnu = u^(n+1)/(n+1)^2[-1+(n+1)ln u] + c

confidence assessment: 3

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16:57:01

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

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u = ln x, du = 1/x dx, dv = x^4 dx, v = x^5/5

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

x^5/5 ln(x) - 1/5 int.(x^4)dx

x^5/5 ln(x) - 1/5(x^5/5) + C

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

confidence assessment: 3

Good.

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16:57:12

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

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This problem is not in the book. Problem# 65 is similar.

confidence assessment:

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16:57:23

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

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.3976

confidence assessment: 2

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16:57:33

Explain how you obtained your result.

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I used substitution to solve. Also use the average cost formula. 1/b-a int f(x)

(.04t - .3)^1/2

u = .04t - .3, du/dx = .04, du = .04 dx, dx = 1/.04 du

1/13-10 (int (.04t - .3)^1/2)

1/3 (1/.04 int u^1/2 du)

1/3 [(2/.12)(.04t - .3)^3/2]13,10

= .3976

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. *&*&

** 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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16:57:36

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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confidence assessment:

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Good work. See my notes.

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Let me know if you have questions. &#