Assignment 5

course Mth 174

|§ÎðªàÞÝÁæl¼¢òíÌþ¥¥Ÿe®ƒassignment #005

¹šØ¯ÿ”`˜âSû‡«¬B©ÀÂàÄÚø

Physics II

06-07-2009

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

11:08:05

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

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

RESPONSE -->

I thought this integral resembled III.14 in the table so I transformed part of it into a polynomial.

I transformed x^4= (x + x^3) so

integral of x^4 e^(3x)= integral of (x + x^3) e^(3x)

= 1/3 [ (x + x^3) e^(ex) - 1/9(1 + 3x^2)e^(3x) + 1/27 (6x) - 1/81 (6)] + C

The integral is of x^4 e^(3 x).

x^4 is a polynomial, and e^(3 x) is of the form e^(a x). So the integrand is of the form

p(x) e^(a x)

with p(x) = x^4 and a = 3.

The correct formula to use is #14

We obtain

p ' (x) = 4 x^3

p '' (x) = 12 x^2

p ''' (x) = 24 x

p '''' (x) = 24.

Thus the solution is

1 / a * p(x) e^(a x) - 1 / a^2 * p ' (x) e^(a x) + 1 / a^3 * p''(x) e^(a x) - 1 / a^4 * p ''' (x) e^(a x) + 1 / a^5 * p''''(x) e^(a x)

= 1 / 3 * x^4 e^(3 x) - 1 / 3^2 * 4 x^3 e^(3 x) + 1 / 3^3 * 12 x^2 e^(3 x) - 1 / 3^4 * 24 x e^(3 x) + 1 / 3^5 * 24 e^(3 x)

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

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

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

11:08:11

what it is your antiderivative?

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

RESPONSE -->

I thought this integral resembled III.14 in the table so I transformed part of it into a polynomial.

I transformed x^4= (x + x^3) so

integral of x^4 e^(3x)= integral of (x + x^3) e^(3x)

= 1/3 [ (x + x^3) e^(ex) - 1/9(1 + 3x^2)e^(3x) + 1/27 (6x) - 1/81 (6)] + C

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

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

11:08:22

Which formula from the table did you use?

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

RESPONSE -->

III.14

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

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

11:09:33

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)?

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

RESPONSE -->

a = 3

p(x) = x + x^3

p '(x) = 1 + 3x^2

p ''(x)= 6x

p '''(x)= 6

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

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

11:26:32

Query problem 7.3.33 (3d edition #30) 1 / [ 1 + (z+2)^2 ) ]

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

RESPONSE -->

This problem looks similar to V.26, so I transformed it:

1 + (z+2)^2= 1 + z^2 + 4z + 4 + 1=

z^2 + 4z + 5= -(z + 5)(z - 1) so

1 / [ 1 + (z+2)^2 ) ] = 1/ [-(z + 5)(z - 1)] = -1/4 (lnI z - 5I - lnIz + 1I + C

a= -5

b = -1

If you let u = x+2 then du = dz and the integrand becomes 1 / (1 + u^2). This is the derivative of arctan(u), so letting u = z+2 gives us the correct result

arctan(z+2) + C

Applying the formula:

z is the variable of integration in the given problem, x is the variable of integration in the table. a is a constant, so a won't be z + 2.

 

By formula 24 the antiderivative of 1 / (a^2 + x^2) is 1/a * arcTan(x/a).

 

Unlike some formulas in the table, this formula is easy to figure out:

 

1 / (a^2 + x^2) = 1 / (a^2( 1 + x^2/a^2) ) = 1/a^2 ( 1 / (1 + (x/a)^2).

 

Let u = x / a so du = dx / a; you get integrand 1 / a * (1 / (1 + u^2) ) , which has antiderivative 1/a * arcTan(u) = 1/a * arcTan(x/a).

 

You don't really need to know all that, but it should clarify what is constant and what is variable.

 

Starting with int(1/ (1+(z+2)^2) dz ), let x = z + 2 so dx = dz. You get

 

int(1/ (1+x^2) dx ), which is formula 24 with a = 1. The result is

 

1/1 * arcTan(x/1), or just arcTan(x). Since x = z + 2, the final form of the integral is

 

arcTan(z+2).

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

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

11:26:37

What is your integral?

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

RESPONSE -->

This problem looks similar to V.26, so I transformed it:

1 + (z+2)^2= 1 + z^2 + 4z + 4 + 1=

z^2 + 4z + 5= -(z + 5)(z - 1) so

1 / [ 1 + (z+2)^2 ) ] = 1/ [-(z + 5)(z - 1)] = -1/4 (lnI z - 5I - lnIz + 1I + C

a= -5

b = -1

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

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

11:26:56

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

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

RESPONSE -->

V.26 This problem looks similar to V.26, so I transformed it:

1 + (z+2)^2= 1 + z^2 + 4z + 4 + 1=

z^2 + 4z + 5= -(z + 5)(z - 1) so

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

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

13:17:19

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

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

RESPONSE -->

I have had some trouble figuring out this section. I couldn't figure out how to break down this denominator to make a

partial fraction. Also I do not understand a step in the process in general. In the #10 class notes it explains another

problem: This one is 1/[(x -3)(x +5)]. I understand this part of the notes:

""We know that when a fraction with denominator x-3 is added to a fraction with denominator x+5 we will obtain a

fraction whose denominator is (x-3)(x+5).

We conclude that it must be possible to express the given fraction as the sum A / (x-3) + B / (x+5), where A and B are

numbers to be determined.

Setting the original fraction equal to the sum we obtain an equation for A and B, as expressed in the third line.

Multiplying both sides of the equation by (x-3) (x+5) we obtain the equation in the fourth line, which we rearrange to the

form in the fifth line by collecting the x terms and the constant terms on the left-hand side and factoring."

This part I do not understand:

""Since the right-hand side does not have an x term, we see that A + B = 0"

How did you find that this equals 0?

The equation for this function would be

A / (x-3) + B / (x+5) = 1/[(x -3)(x +5)]

To simplify the left-hand side need to obtain a common denominator. We multiply the first term by (x + 5) / (x + 5), and the second term by (x - 3) / (x - 3):

A / (x-3) * (x + 5) / (x + 5) + B / (x+5) * (x - 3) / (x - 3) = 1/[(x -3)(x +5)] so that

A ( x + 5) / ( (x - 3) ( x + 5) ) + B ( x - 3) / ( (x - 3) (x + 5) ) = 1/[(x -3)(x +5)] . Adding the fractions on the left-hand side:

( A ( x + 5) + B ( x - 3) ) / ( ( x - 3) ( x + 5) ) = 1/[(x -3)(x +5)] . Simplifying the numerator we have

( (A + B) x + (5 A - 3 B) ) / ( ( x - 3) ( x + 5) ) = 1/[(x -3)(x +5)]. The denominators are equal, so the equation is solved if the numerators are equal:

(A + B) x + (5 A - 3 B) = 1.

It is this last equation which lacks an x term on the right-hand side. To maintain equality the left-hand side must also have no x term, which can be so only if A + B = 0.

The other term 5 A - 3 B is equal to 1.

Thus we have the simultaneous equations

A + B = 0

5 A - 3 B = 1.

These equations are easily solve, yielding the solution A = 1/8, B = -1/8.

I understand this:

""we see that therefore 5A - 3B = 1, so we have two equations in two unknowns A and B.""

I could not figure out how you found A and B as shown below:

Solving these equations we obtain B = -1/8, A = 1/8, as indicated.

We conclude that the expression to be integrated is A / (x-3) + B / (x+5) = 1/8 * 1/(x-3) - 1/8 * 1/(x+5).

The system

A + B = 0

5 A - 3 B = 1.

can be solved by elimination or substitution.

Using substitution:

Solve the first equation for A, obtaining A = -B.

Substitute this value of A into the second equation. obtaining

5 * (-B) + (-3 B) = 1

so that

-8 B = 1 and

B = -1/8.

Go back to the fact that A = -B to obtain

A = - (-1/8) = 1/8.

To solve by elimination you could add 3 times the first equation to the second, eliminating B and obtaining

8 A = 1, so that

A = 1/8.

Substituting this back into the first equation we obtain

1/8 + B = 0 so that

B = -1/8.

We conclude that the expression to be integrated is A / (x-3) + B / (x+5) = 1/8 * 1/(x-3) - 1/8 * 1/(x+5).""

For the present problem:

The denominator factors by grouping:

 

 y^3 - y^2 + y – 1 = (y^3 + y) – (y^2 + 1) = y ( y^2 + 1) – 1 ( y^2 + 1) = (y – 1) ( y^2 + 1).

 

Using partial fractions you would then have

 

(a y + b) /(y^2 + 1) + c /(y-1) = y / ( (y^2+1)(y-1)) where we need to evaluate a, b and c.

 

Putting the left-hand side over a common denominator (multiply the first fraction by (y-1)/(y-1) and the second by (y^2+1) / (y^2 + 1)) we obtain:

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

The denominators are identical so the numerators are equal, giving us

 

(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).

 

An 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.

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

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

13:17:22

What is your result?

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

RESPONSE -->

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

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

13:17:26

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

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

RESPONSE -->

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

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

13:17:31

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

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

RESPONSE -->

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

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

18:16:58

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

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

RESPONSE -->

First I complete the square and get

2z - z^2 = 1- (x-1)^2

Now I can substitute z-1= sinz or x = sinz + 1 and dz = cosz dz

Integral (z-1)/`sqrt(2z-z^2)dz= integral (z-1)/ sqrt 1- (z - 1)^2 = [(sin(theta)/ (sqrt 1 - sin^2(theta)) ] cos(theta) d(theta)

= integral [sin(theta)/ cos(theta)] cos(theta) d(theta)

= integral sin(theta) d(theta)

= integral sin(theta) + C

Since z-1 = sin(theta), theta=arcsin(x - 1) + C

 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).

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

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

18:17:19

What did you get for your integral?

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

RESPONSE -->

arcsin(x - 1) + C

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

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

18:18:14

What substitution did you use?

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

RESPONSE -->

First I complete the square and get

2z - z^2 = 1- (x-1)^2

Then I can substitute z-1= sinz or x = sinz + 1 and dz = cosz dz

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

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

18:41:59

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

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

RESPONSE -->

First complete the square so

(2y^2 + 3y + 1) = (2y + .5)^2 + (3/4)= (2y + .5)^2 + (sqrt 3/2)^2

x + .5= (sqrt3/2)tan(theta) dx= (sqrt3/2)(1/cos^2(theta) d(theta) so

integral (y+2) / (2y^2 + 3y + 1)=integral [(y + 2)/ (2y + .5)^2 + (3/4)] * [ (sqrt 3/2) (1/cos^2(theta)] dtheta

=(sqrt 3/2) integal (y + 2)/ ((3/4tan^29(theta) + 3/4))(1/cos^2(theta)) d(theta)

= 2/sqrt 3 integral (y + 2)/ (tan^2(theta) + 1)(cos^2(theta) d(theta)

= 2/sqrt 3 integral ( y + 2)/ in^2(theta) + cos^2(theta) d(theta)

= 2/sqrt 3 integral (y + 2) d(theta)

= (2/sqrt 3 )* y^2/2 + 2y(theta) + C

theta = arctan((2/sqrt 3) + (1/sqrt3))

=(2/sqrt 3) * y^2/2 + 2y arctan((2/sqrt 3) + (1/sqrt3)) + C

(y+2) / (2y^2 + 3 y + 1) =

(y + 2) / ( (2y + 1) ( y + 1) ) =

(y + 2) / ( 2(y + 1/2) ( y + 1) ) =

1/2 * (y + 2) / ( (y + 1/2) ( y + 1) )

The expression

(y + 2) / ( (y + 1/2) ( y + 1) )

is of the form

(cx + d) / ( (x - a)(x - b) )

with c = 1, d = 2, a = -1/2 and b = -1.

Its antiderivative is given as

1 / (a - b) [ (ac + d) ln | x - a | - (bc + d) ln | x - b | ] + C.

The final result is obtained by substitution.

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

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

18:42:02

What is your integral and how did you obtain it?

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

RESPONSE -->

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

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

18:42:54

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

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

RESPONSE -->

I'm still pretty confused with the section. I'm going to read over it again and try a few more problems and send you my questions. Thanks!

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

&#Good work. See my notes and let me know if you have questions. &#