Asst 4

course Mth 174

06-25-2007......!!!!!!!!...................................

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

15:22:20

what is the requested antiderivative?

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antiderivative of sin^2 x

u = sin x v = -cos x

u' = cos x v' = sin x

-sinx cosx + int(cos^2 x)

-sinx cosx + int(1-sin^2 x)

-sinx cosx + int(1) - int (sin^2 x)

2 int (sin^2 x) dx = -sin x cos x + int(1) dx

-1/2 sin x cos x + 1/2 x + C

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15:23:16

What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?

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antiderivative of sin^2 x

I used u & v...

u = sin x v = -cos x

u' = cos x v' = sin x

Then, cos^2 x = 1 - sin^2 x, so I plugged that in for the cos^2 x...

-sinx cosx + int(cos^2 x)

-sinx cosx + int(1-sin^2 x)

-sinx cosx + int(1) - int (sin^2 x)

2 int (sin^2 x) dx = -sin x cos x + int(1) dx

-1/2 sin x cos x + 1/2 x + C

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15:32:35

what is the requested antiderivative?

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RESPONSE -->

(t+2) `sqrt(2+3t)

let u = t +1 and v = 3/6 (2+3t)^(3/2)

u'= 1 v'= (2+3t)^(1/2)

int [(t+2) `sqrt(2+3t)] =

3/6 (t+1) (2+3t)^(3/2) - int[1* 3/6 (2+3t)^(3/2)]

3/6 (t+1) (2+3t)^(3/2) - (3/6) int[(2+3t)^(3/2)]

add one to 3/2 = 5/2

divide by 5/2

giving 3/6 * 2/5 = 6/30 = 1/5

3/6 (t+1) (2+3t)^(3/2) - 1/5 (2+3t)^(5/2)

If you use

u=t+2
u'=1
v'=(2+3t)^(1/2)
v=2/9 (3t+2)^(3/2)

then you get

2/9 (t+2) (3t+2)^(3/2) - integral( 2/9 (3t+2)^(3/2) dt ) or
2/9 (t+2) (3t+2)^(3/2) - 2 / (3 * 5/2 * 9) (3t+2)^(5/2) or
2/9 (t+2) (3t+2)^(3/2) - 4/135 (3t+2)^(5/2). Factoring out (3t + 2)^(3/2) you get

(3t+2)^(3/2) [ 2/9 (t+2) - 4/135 (3t+2) ] or
(3t+2)^(3/2) [ 30/135 (t+2) - 4/135 (3t+2) ] or
(3t+2)^(3/2) [ 30 (t+2) - 4(3t+2) ] / 135 which simplifies to

2( 9t + 26) ( 3t+2)^(3/2) / 135.

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15:32:45

What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?

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RESPONSE -->

(t+2) `sqrt(2+3t)

let u = t +1 and v = 3/6 (2+3t)^(3/2)

u'= 1 v'= (2+3t)^(1/2)

int [(t+2) `sqrt(2+3t)] =

3/6 (t+1) (2+3t)^(3/2) - int[1* 3/6 (2+3t)^(3/2)]

3/6 (t+1) (2+3t)^(3/2) - (3/6) int[(2+3t)^(3/2)]

add one to 3/2 = 5/2

divide by 5/2

giving 3/6 * 2/5 = 6/30 = 1/5

3/6 (t+1) (2+3t)^(3/2) - 1/5 (2+3t)^(5/2)

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15:32:49

What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?

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RESPONSE -->

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15:35:17

query problem 7.2.27 antiderivative of x^5 cos(x^3)

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16:03:57

what is the requested antiderivative?

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int[x^5 cos(x^3)]

let u = cos(x^3) and v = 1/6 x^6

u'= -3x^2sin(x^3) v' = x^5

1/3 x^3sin(x^3) - int [-3x^2 sin(x^3) * 1/6 x^6]

1/3 x^3sin(x^3) - 1/6 int [[-3x^2 sin(x^3) * 1/6 x^6]

Here is the problem I run into every time...what to do with the [-3x^2 sin(x^3) * 1/6 x^6] - I don't get the answers to the questions to ever appear and I can't figure out why so I don't knwo what to do. Please help.

The integral you get from your breakdown is at least as difficult to integrate as the original form, so your breakdown did not lead to progress, as you noted. You have to try different breakdowns.

In this case you want to ask yourself what different powers of x could be used for u. With some trial and error this approach might lead you to the following solution:

Let u = x^3, v' = x^2 cos(x^3).

Then u' = 3 x^2 and v = 1/3 sin(x^3) so you have

1/3 * x^3 sin(x^3) - 1/3 * int(3 x^2 sin(x^3) dx).

Now let u = x^3 so du/dx = 3x^2. You get

1/3 * x^3 sin(x^3) - 1/3 * int( sin u du ) = 1/3 (x^3 sin(x^3) + cos u ) = 1/3 ( x^3 sin(x^3) + cos(x^3) ).

It's pretty neat the way this one works out, but you have to try a lot of u and v combinations before you stumble on the right one.

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16:06:24

What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?

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int x^5 cos(x^3)

I got confused - but I tried

u = x ^ 5 v = 1/(3x^2)sin(x^3

u' = 5x4 v' =cos(x^3)

and

u = cos(x^3) v = 1/6 x^6

u'= -3x^2 sin(x^3) v' = x^5

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16:07:44

06-25-2007 16:07:44

query problem 7.2.52 (3d edition #50) f(0)=6, f(1) = 5, f'(1) = 2; find int( x f'', x, 0, 1).

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NOTES -------> f(0)=6, f(1) = 5, f'(1) = 2; find int( x f'', x, 0, 1).

I'm not even sure where to begin with this and again - I cannot make the answers come up so I'm not sure how to help myself. Sorry.

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16:08:15

What is the value of the requested integral?

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again I'm not sure where to start - and the answers don't come up when after I click enter response then next question/answer.

You can find this one using integration by parts:

Let u=x and v' = f''(x). Then

u'=1 and v=f'(x).

uv-integral of u'v is thus

xf'(x)-integral of f'(x)

Integral of f'(x) is f(x). So antiderivative is

x f ' (x)-f(x), which we evaluate between 0 and 1. Using the given values we get

1 * f'(1)- (f(1) - f(0)) =
f ‘ (1) + f(0) – f(1) =
2 + 6 - 5 = 3.

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16:08:31

How did you use integration by parts to obtain this result? Be specific.

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again I didn't get this and the answers don't come up.

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16:09:09

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

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I tried again in another Assignment and still the answers didn't come up.

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16:09:13

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

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assignment #004

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Physics II

06-25-2007

The Queries for this course do not include the answers, which will be inserted in the manner shown here when I post your work.

You got some of these, and you know how to use the rules for basic problems. You just need practice so you can handle the trickier problems. The solutions given here should be helpful.

&#You are always welcome to ask self-contained questions about anything. By self-contained questions I mean a question that includes a brief statement of the problem or topic you are asking about (in order to give everyone the best responses I can, I can't take time to look problems up in the text, which I don't carry with me in any case), and a statement of precisely what you do and do not understand about the situation. If it's a problem, you should include a list of things you have tried in attempting to solve (or to understand) the problem. Depending on the problem this might include a description of any diagrams, listings of concepts and topics you think might be helpful, and other relevant information. This can be relatively brief, but the more you can tell me, the more you will learn in the process, and the more specifically I can address my response. &#