Assignment 19

course Mth 272

Ճ×߉†å¦m»—ª¹Ö°¿—„Õü°®WbæÑassignment #019

019.

Applied Calculus II

03-16-2009

......!!!!!!!!...................................

01:10:50

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

......!!!!!!!!...................................

RESPONSE -->

i used 41. in the table

x^2+1/(2+1)^2 [-1+(3+1) ln x^2] +c

x^3/ 9 [3lnx^2] + c

confidence assessment: 3

Some of your simplification seems to contradict the order of operations, and I don't think #41 applies to this integral.

In any case:

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.

.................................................

......!!!!!!!!...................................

01:10:56

What is your result?

......!!!!!!!!...................................

RESPONSE -->

answered

confidence assessment: 3

.................................................

......!!!!!!!!...................................

01:11:35

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

......!!!!!!!!...................................

RESPONSE -->

I used #41 and i just plugged in the necessary values

confidence assessment: 3

.................................................

......!!!!!!!!...................................

01:17:51

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

......!!!!!!!!...................................

RESPONSE -->

i used #41 again

x^5/ (25) [5lnx]

confidence assessment: 3

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)

.................................................

......!!!!!!!!...................................

01:17:56

What is your integral?

......!!!!!!!!...................................

RESPONSE -->

confidence assessment:

.................................................

......!!!!!!!!...................................

01:18:11

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

......!!!!!!!!...................................

RESPONSE -->

# 41 again, by plugging in the values

confidence assessment: 3

.................................................

......!!!!!!!!...................................

01:18:49

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

......!!!!!!!!...................................

RESPONSE -->

you just find you u and du and then you can plug them in as needed

confidence assessment: 3

.................................................

......!!!!!!!!...................................

01:32:56

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

......!!!!!!!!...................................

RESPONSE -->

finding the integral using the table

= 1/2(375.6 t^2*sqrt((375.6 t^2- 715.86) - 715.86 ln [375.6 t^2 + sqrt(375.6 t^2+ 715.86))

from [8, 16]

1/2(375.6 (16)^2*sqrt((375.6 t^2- 715.86) - 715.86 ln [375.6 t^2 + sqrt(375.6 t^2+ 715.86))

..replace w/ 16

minus

1/2(375.6 t^2*sqrt((375.6 t^2- 715.86) - 715.86 ln [375.6 t^2 + sqrt(375.6 t^2+ 715.86))

... replace with 8

= 29696515.01 - 1828312.603 = 1141202.407 = P

confidence assessment: 3

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.

.................................................

......!!!!!!!!...................................

01:33:05

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

......!!!!!!!!...................................

RESPONSE -->

answered

confidence assessment: 3

.................................................

......!!!!!!!!...................................

01:33:14

Explain how you obtained your result.

......!!!!!!!!...................................

RESPONSE -->

answered

confidence assessment: 3

.................................................

......!!!!!!!!...................................

01:33:33

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

......!!!!!!!!...................................

RESPONSE -->

I couldn't find the last problem in the book.

confidence assessment: 3

.................................................

You probably oversimplified a couple of these problems, using a formula that doesn't apply to all of them. See my notes and let me know if you have questions.