Assignments 4  5

course Mth 174

YЫkassignment #004

ʣ֗ngymӮg]~w

Physics II

09-25-2007

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17:37:25

query problem 7.2.12 (3d edition #11) antiderivative of sin^2 x

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RESPONSE -->

the antiderivative of sin^2x is -cos^3 / 3

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17:37:51

what is the requested antiderivative?

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RESPONSE -->

the requested antiderivative is -cos^3 / 3

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17:38:25

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

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RESPONSE -->

i didnt use the substitution method but i did use the power rule and the sin/cos rule to find the antiderivative.

cos^3 / 3 isn't even a function, having no argument.

You probably meant cos^3(2x) / 3, but the derivative of cos^3(2x) is -6 sin(2x) cos^2(2x); this derivative requires two applications of the chain rule.

The only way you are likely to find a correct antiderivative is to use appropriate techniques. In this case you could use integration by parts or trigonometric identities such as sin^2(x) = (1 - cos(2x) ) / 2 and sin(2x) = 2 sin x cos x.

Solution by trigonometric identities:

sin^2(x) = (1 - cos(2x) ) / 2 so the antiderivative is

1/2 ( x - sin(2x) / 2 ) + c =

1/2 ( x - sin x cos x) + c;

note that sin(2x) = 2 sin x cos x

Integration by parts:

Let u= sinx and dv = sinx dx. Then v = -cos(x) and

u v int(v du) = -sinx cosx + int (cos^2 x)

= -1/2 sinx cosx + 1/2 x + C

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17:42:06

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

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RESPONSE -->

take the square and more it to a power which would make it (t+2) (2+3t)^(1/2).

so the antiderivative is (t^2+2t) (3/2)(2t+3t^2)^(3/2)

The derivative of your result is not the original function.

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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17:42:34

what is the requested antiderivative?

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RESPONSE -->

the requested antiderivative is the one that i just wrote in the question before but you add a +C on the end of it.

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17:43:02

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

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RESPONSE -->

i used the power rule, the product rule, and i put the square root rule into play.

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17:44:39

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

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RESPONSE -->

(1/3)x^3 sinx^3 + (1/3) cosx^3 is the antiderivative that is associate with this problem

That is correct, but that result could have been obtained without using appropriate integration technique so the analytical process must be documented:

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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17:44:43

what is the requested antiderivative?

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RESPONSE -->

(1/3)x^3 sinx^3 + (1/3) cosx^3 + C

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17:45:09

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

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RESPONSE -->

i used the power rule of course again, used the product rule and the sin/cos rule.

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17:54:14

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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RESPONSE -->

the integral in this situation is int(x f'', x, 0, 1) = x f'' x I(0,1)dx - int (x f'', x) dx

= xf'' x (0,1) - int(2xf'' 0,1)

= x 2 x - 2x int (0,1)

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

= 0

Compare with the following:

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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17:54:18

What is the value of the requested integral?

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RESPONSE -->

the integral in this situation is int(x f'', x, 0, 1) = x f'' x I(0,1)dx - int (x f'', x) dx

= xf'' x (0,1) - int(2xf'' 0,1)

= x 2 x - 2x int (0,1)

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

= 0

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17:54:30

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

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RESPONSE -->

the integral in this situation is int(x f'', x, 0, 1) = x f'' x I(0,1)dx - int (x f'', x) dx

= xf'' x (0,1) - int(2xf'' 0,1)

= x 2 x - 2x int (0,1)

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

= 0

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17:55:12

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

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RESPONSE -->

i did not totally understand that integration by parts thing using the formula used in the book. I got use to the formula though and by the end of the assignment i the formula was really easy to use.

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17:55:20

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

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RESPONSE -->

already did

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Vӂb

assignment #005

ʣ֗ngymӮg]~w

Physics II

09-25-2007

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17:56:26

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RESPONSE -->

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17:57:03

Query problem 7.3.17 (3d edition #15) x^4 e^(3x)

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RESPONSE -->

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17:58:29

what it is your antiderivative?

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RESPONSE -->

the antiderivative that i got was

((1/3)x^4 - (4/9)x^3 + (4/9)x^2 - (8/27)x + (8/81))e^3x

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18:00:42

Which formula from the table did you use?

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RESPONSE -->

i used the formula III-14

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18:02:51

You should have used formula 14. What was your value of a, what was p(x), how many derivatives did you use and what were your derivatives of p(x)?

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RESPONSE -->

yes i used the formula 14 and the value of my a was (1/3) and the p(x) value of it was x^4. i had to take 5 derivatives and their values of p(x) were x^3, x^2, x, and 1.

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18:03:41

What is your integral?

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RESPONSE -->

the integral that i got was

arctan (z+2) + C

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18:05:52

Which formula from the table did you use and how did you get the integrand into the form of this formula?

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RESPONSE -->

i used the formula #24 and to get the integrand into the form of this formula you have to flip the integral and then divide by 1.

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18:15:27

7.4.6 (#8 in 3d edition). Integrate 2y / ( y^3 - y^2 + y - 1)

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RESPONSE -->

the integration here is using the long division method breaking the functions up into partial fractions which are

2y / (y^2+1) (y-1)

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18:15:32

What is your result?

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RESPONSE -->

2y / (y^2+1) (y-1)

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18:16:18

How did you factor your denominator to get the integrand into a form amenable to partial fractions?

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RESPONSE -->

i factored it by grouping the y^3 and y^2 together and factored out a y^2 and then i grouped the y and 1 together and just used it as one of my partial fractions.

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18:16:42

After the integrand was in the form 2y / [ (y-1) * (y^2 + 1) ], what form of partial fraction did you use?

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RESPONSE -->

i used the form of long division

Using partial fractions you would have

(a y + b) /(y^2 + 1) + c /(y-1) = y / ( (y^2+1)(y-1)) so that

[ (a y + b)(y-1) + c(y^2+1) ] / ((y^2+1)(y-1)) = y / ((y^2+1)(y-1)). Thus

(a y + b)(y-1) + c(y^2+1) = y, or

a y^2 + (-a + b) y - b + c y^2 + c = y. Grouping the left-hand side:

(a + c) y^2 + (a - b) y + c - b = y. Since this must be so for all y, we have

a + c = 0 (this since the coefficient of y^2 on the right is 0)

-a + b = 1 (since the coefficient of y on the right is 1)

c - b = 0 (since there is no constant term on the right).

From the third equation we have b = c; from the first a = -c. So the second equation

becomes

c + c = 1, giving us 2 c = 1 and c = 1/2.

Thus b = c = 1/2 and a = -c = -1/2.

Our integrand (a y + b) /(y^2 + 1) + c /(y-1) becomes

(-1/2 y + 1/2 ) / (y^2 + 1) + 1/2 * 1 / (y-1), or

-1/2 y / (y^2 + 1) + 1/2 * 1 / (y^2 + 1) + 1/2 * 1 / (y-1).

The antiderivative is easily enough found with or without tables to be

-1/2 ln(y^2 + 1) + 1/2 arcTan(y) + 1/2 ln | y - 1 | + c.

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18:18:16

7.4.29 (3d edition #24). Integrate (z-1)/`sqrt(2z-z^2)

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RESPONSE -->

to integral of this problem is

w = 1-(z-1)^2

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18:18:34

What did you get for your integral?

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RESPONSE -->

w = 1-(z-1)^2

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18:18:47

What substitution did you use?

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RESPONSE -->

i used the substitution of u and du

If you let u = 2z - z^2 you get du = (2 - 2z) dz or -2(z-1) dz; thus (z-1) dz = -du / 2.

So (z-1) / `sqrt(2z - z^2) dz = 1 / sqrt(u) * (-du/2) = - .5 u^-.5 du, which integrates to

-u^.5. Translated in terms of the original variable z we get

-sqrt(2z-z^2).

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18:22:59

7.4.40 (3d edition #28). integrate (y+2) / (2y^2 + 3y + 1)

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RESPONSE -->

the integral here is y +2 / (2y^2 + (1/2))^2 + (square3 / 2)^2

(y + 2) / ( (2y + 1)(y + 1) ) = A /(2y + 1) + B / (y + 1) = [ A(y+1) + B(2y + 1) ] / ( (2y + 1)(y + 1) ) so

A(y+1) + B(2y+1) = y + 2 or

(A + 2B) y + (A + B) = y + 2. Thus

A + 2B = 1 and

A + B = 2 so

A + 2( 2 - A ) = 1 or

-A = -3 and A = 3.

Thus B = -1.

The function is 3 / (2y + 1) - 1 / (y + 1) and its antiderivative is 3/2 ln | 2y + 1 | - ln | y + 1 |.

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18:23:22

What is your integral and how did you obtain it?

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RESPONSE -->

it is the response i had for the question before and i obtained it by using the completing the square method

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18:23:46

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

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RESPONSE -->

i didnt really understand the completing the square things in the 7.4 section.

I'll be glad to start a dialog with you on the process. If you wish send me a copy of the following question along with your answer:

The square of (a + b) is a^2 + 2 a b + b^2. Are you familiar with this formula and do you see how to get it using the distributive law?

Can you recognize which of the following four expressions is/are perfect square(s)?

(x^2 - 16), (x^2 + 8 x + 16), (x^2 + 16), (x^2 + 12 x + 16)

Can you complete the square on x^2 + 6?

Specifically, what number could be added to x^2 + 6 to get a perfect square?

Answer these questions and I'll send you another series to take the process a little further.

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׆l笄ʨ

assignment #005

ʣ֗ngymӮg]~w

Physics II

09-25-2007

To see whether you have correctly integrated an expression, take the derivative of your result and see if it is equal to the original expression.

You generally need to use the techniques shown in my responses to find the antiderivatives. In most cases there is no easier way and no shortcut.