Assignment 4 Query

course Mth 174

Yi| ܟݝassignment #004

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Physics II

09-21-2007

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20:47:46

what is the requested antiderivative?

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RESPONSE -->

I found the antiderivative to be:

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

I found this by going through integration by parts:

int(sin^2 x)

u = sin x u' = cos x

v' = sin x v = -cos x

int(sin^2 x)= -cosx sinx - int(-cos^2 x)

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

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

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

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

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

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20:47:58

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

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RESPONSE -->

Whoops. See prior answer.

It's always fine to give the details before they are requested.

Good solution.

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20:54:01

query problem 7.2.16 antiderivative of (t+2) `sqrt(2+3t)

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RESPONSE -->

I found the antiderivative to be:

= (2/9)(t + 2)(2 + 3t)^(3/2) - (4/45)(2 + 3t)^(5/2) + C

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20:54:08

what is the requested antiderivative?

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RESPONSE -->

See prior answer.

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21:00:57

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

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RESPONSE -->

First I split it up into u and v:

u = (t + 2) u' = 1

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

Using copy and paste:

int( (t + 2)(2 + 3t)^(1/2) ) = (t + 2)(2/9)(2 + 3t)^(3/2) - int( (2/9)(2 + 3t)^(3/2) )

= (t + 2)(2/9)(2 + 3t)^(3/2) - (2/9)(2/5)(2 + 3t)^(5/2)

= (t + 2)(2/9)(2 + 3t)^(3/2) - (2/45)(2 + 3t)^(5/2)

in your second term, 2/9 * 2/5 = 4/45; however there's also a factor of 3 due to the 3t. Otherwise very good. Check out the details and also the final simplification below:

You have

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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21:01:20

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

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RESPONSE -->

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21:03:00

what is the requested antiderivative?

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RESPONSE -->

I found the antiderivative to be:

(1/3) x^3 sin x^3 + (1/3) cos x^3 + C

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21:08:27

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

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RESPONSE -->

Tricks. I like that. There sure are a lot of tricks to remember, especially that one with the big C...

Anyway, here is what I did:

int( x^5 cos x^3 )

u = x^3 u' = 3x^2

v' = x^2 cos x^3 v = (1/3)sin x^3

int = x^3(1/3)sin x^3 - int( x^2 sin x^3 )

= x^3(1/3)sin x^3 - ( -(1/3)cos x^3 )

= (1/3) x^3 sin x^3 + (1/3)cos x^3 + C

Very good.

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The problem was not included in your SEND file:

**** query problem 7.2.50 was 7.3.48 f(0)=6, f(1) = 5, f'(1) = 2; find int( x f'', x, 0, 1).

21:29:44

What is the value of the requested integral?

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RESPONSE -->

I found the value of the requested integral to be:

3.

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21:39:04

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

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RESPONSE -->

Ok, here we go:

First find the general antiderivative:

int( x f''(x) dx )

u = x u' = 1

v' = f''(x) v = f'(x)

int = x f'(x) - int( f'(x) dx )

= x f'(x) - f(x)

So now I can find the specific integral by using the formula F(b) - F(a):

1 f'(1) - f(1) - ( 0 f'(0) - f(0) )

= 1 * 2 - 5 - ( 0 - 6)

= 2 - 5 + 6

= 3

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21:41:08

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

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RESPONSE -->

That last one was surprisingly cool. Once I figured out that I didn't need to worry about the value of the derivatives at points other than those give, it was simple. It was also very expressive of what I have done before so many times, but in a generic sense rather than a specific one.

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21:51:49

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

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Um, I think I did. More? Okay. Integration by parts can take a while (and be a little tricky getting started), but the ending is fairly simple to get to. I would hate to have to reverse the product rule without a simple series of steps like these to follow.

Ah, I do have one question. In 7.2.35, here is the integral:

int( ln(1 + t) dt, 0, 5, t)

I selected these values of u and v', and found u' and v:

u = ln(1 + t) u' = 1 / (1 + t)

v' = 1 v = t

But it seems that I was incorrect in v, since the book's answer is v = (1 + t). Is there any way to figure this out without going through the whole substition by parts process and reverse-checking?

Is there a connection between ln(1 + t) and (1 + t) that will stick around in other problems?

I guess that covers it. Thanks!

If you let x = 1 + t the du = dx and the integral converts to just int(ln(x) dx).

To integrate ln(x) let u = ln(x) and v ' = 1 so that u ' = 1/x and v = x. You get

x ln x - int(1/x * x dx) = x ln x - x; since x = 1 + t this is

(1 + t) ln(1 + t) - (1 + t).

However when doing the definite integral I would just convert the t limits 0 to 5 into x limits 1 and 6, and use the x ln x - x form.

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Very good work. Be sure to see my notes and let me know if you have additional questions.