assignment 14

course mth 272

I found this assignment difficult.

......!!!!!!!!...................................Applied Calculus II

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Asst # 14

07-05-2006

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17:09:25

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**** Query problem 6.1.5 (was 6.1.4) integral of (2t-1)/(t^2-t+2)

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17:09:47

ln (t^2-t+2) + C

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17:09:48

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**** What is your result?

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17:10:00

ln ( t^2-t+2) + C

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17:10:01

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**** What substitution did you use and what was the integral after substitution?

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17:10:41

general log rule: (du/dx) / u dx = 1/u = ln (u) + C

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17:10:42

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**** Query problem 6.1.13 (was 6.1.32) integral of 1 / (`sqrt(x) + 1)

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17:10:57

2 (x+1)^(1/2) + C

Your answer would have been correct if the expression was 1 / sqrt(x + 1), but I believe the expression is 1 / (sqrt(x) + 1), which is different and leads to a very different result.

If we let u = (sqrt x) + 1 we can solve for x to get

x = (u-1)^2 so that
dx = 2(u-1) du.

So the integral of 1 / (`sqrt(x) + 1) dx becomes the integral of (2 ( u - 1 ) / u) du.

Integrating ( u - 1) / u with respect to u we express this as

( u - 1) / u = (u / u) - (1 / u) = 1 - (1 / u). An antiderivative is u - ln | u |.

Substituting u = (sqrt x) + 1 and adding the integration constant c we end up with

(sqrt x) + 1 - ln | (sqrt x) + 1 | + c.

Our integral is of 2 (u-1)/u, double the expression we just integrated, so our result will also be double. We get

2 (sqrt x) + 2 - 2 ln | (sqrt x) + 1 | + c.
Since c is an arbitrary constant, 2 + c is also an arbitrary constant so the final solution can be expressed as
2 (sqrt x) - 2 ln | (sqrt x) + 1 | + c.

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17:10:59

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**** What is your result?

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17:11:15

2 ( x+1) ^ (1/2) + C

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17:11:16

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**** What substitution did you use and what was the integral after substitution?

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17:13:54

general power rule: u^n+1 / n+1 + C

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17:13:55

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**** If you let u = `sqrt(x) + 1 then what is du and, in terms of u, what is x?

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17:15:57

du = 1

x = u^2-1

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17:15:58

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**** If you solve du = 1 / (2 `sqrt(x)) dx for dx, then substitute your expression for x in terms of u, what do you finally get for dx?

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17:18:29

dx = du( 2 sqrt (x) and u*2-1 = x so

u^2-1 - du(2 sprt (x)) = dx

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17:18:30

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**** What therefore will be your integrand in terms of u and what will be your result, in terms of u?

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17:20:11

u = sqrt.(du(2sprt(x)) dx -1)

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17:20:12

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**** What do you get when you substitute `sqrt(x) + 1 for u into this final expression?

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17:21:02

x= du(2sqt (x) dx - 2

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17:21:03

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**** query problem 6.1.56 (was 6.1.46) area bounded by x (1-x)^(1/3) and y = 0

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17:29:19

u = (1-x) ^ (1/3)

u^3= 1-x

x= 1-u^3

dx=-3u^2du

not sure what to do next or if identifying these is how to start.

Begin by sketching a graph to see that there is a finite region bounded by the graph and by y = 0. Then find the points where the graph crosses the line y = 0 but finding where the expression takes the value 0:

** x(1-x)^(1/3) = 0 when x = 0 or when 1-x = 0. Thus the graph intersects the x axis at x = 0 and x = 1.

We therefore integrate x (1-x)^(1/3) from 0 to 1.

We let u = 1-x so du = dx, and x = 1 - u.

This transforms the integrand to (1 - u) * u^(1/3) = u^(1/3) - u^(4/3), which can be integrated term by term, with each term being a power function.

Our antiderivative is 3/4 u^(4/3) - 3/7 u^(7/3), which translates to 3/4 (1-x)^(4/3) - 3/7 ( 1-x)^(7/3).

The result is 9/28 = .321 approx..

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17:29:20

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**** What is the area?

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17:29:25

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17:29:26

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**** What integral did you evaluate to obtain the area?

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17:29:30

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17:29:31

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**** What substitution did you use to evaluate the integral?

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17:29:35

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17:29:35

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**** Query problem 6.1.64 P = int(1155/32 x^3(1-x)^(3/2), x, a, b).

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17:50:28

from 0 to 25: found 1155/32 (1-u^2/3) (u) (-2/3 ^-.33) du between 1 and .91

but from the example in the book i still am not clear how to solve

The probability of an occurrence between 0 and .25 is found by integrating the expression from x = 0 to x = .25:

Let u = 1-x so du = -dx and x = 1-u.

Express in terms of u:

-(1155/32) * int ( (1-u)^3 (u)^(3/2) du )

Expand the integrand:

-(1155/32) * int( (1 - 3 u + 3 u^2 - u^3) * u^(3/2) ) =

-1155/32 * int( u^(3/2) - 3 u^(5/2) + 3 u^(7/2) - u^(9/2) ) .

An antiderivative of u^(3/2) - 3 u^(5/2) + 3 u^(7/2) - u^(9/2) is 2/5 u^(5/2) - 3 * 2/7 u^(7/2) + 3 * 2/9 * u^(9/2) - 2/11 u^(11/2) . So we obtain for the indefinite integral

-1155/32 * ( 2/5 u^(5/2) - 3 * 2/7 u^(7/2) + 3 * 2/9 * u^(9/2) - 2/11 u^(11/2) ).

Express in terms of x:

-1155/32 * ( 2/5 ( 1 - x )^(5/2) - 3 * 2/7 ( 1 - x )^(7/2) + 3 * 2/9 * ( 1 - x )^(9/2) - 2/11 ( 1 - x )^(11/2) )

Evaluate this antiderivative at the limits of integration 0 and .25 .

You get a probability of .0252, approx, which is about 2.52%, for the integral from 0 to .25; this is the probability of a result between 0 and 25%.

To get the probability of a result between 50% and 100% integrate between .50 and 1. You get .736, which is 73.6%.

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17:50:28

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**** What is the probability that a sample will contain between 0% and 25% iron?

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17:50:44

solve the long expression to get the probability

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17:50:45

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**** What is the probability that a sample will contain between 50% and 100% iron?

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17:51:01

use other limits with the same expression and solve

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17:51:01

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**** What substitution or substitutions did you use to perform the integration?

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17:51:39

constant rule and simple power rule

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17:51:39

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

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17:52:56

this material seems a bit difficult the way they explained in the textbook. other than that, in integration by substitution, it seems like you have to memorize the basic formulas and in turn recognize how to use them.

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You do need to know the basic formulas. Then it takes practice with a large number of problems to learn to recognize the common patterns. Let me know if you have questions.