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.
Question: `qQuery problem 6.4.16 use table to integrate x^2 ( ln(x^3) )^2
Your solution:
Confidence Assessment:
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. </h3>
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·
Self-critique (if necessary):
Self-critique Rating:
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
Your solution:
Confidence Assessment:
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)
Self-critique (if necessary):
Self-critique Rating:
Question: `qQuery problem 6.4.63 profit function P = `sqrt( 375.6 t^2 - 715.86) on [8,16].
(
Your solution:
Confidence Assessment:
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(
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(
Self-critique (if necessary):
Self-critique Rating:
Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.