Query 5

#$&*

course MTH 174

1/28 5:45PM

If your solution to stated problem does not match the given solution, you should self-critique per instructions at 

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

 

Optional preliminary questions:

 

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Question:  `q001.  A typical table of integrals includes the formula

 

integral( c / (ax + b) dx) = c / a * ln | a x + b | + C

 

As with many integrals on typical tables, this integral is easily enough found using standard techniques (in this case just let u = a x + b and proceed in the usual manner).  So while you should not require a table for the present example, or for most of the integrals given in this assignment, it is likely that you will at some point encounter integrals requiring techniques you don't know and need to revert to tables.

 

We will use the formula given above to integrate the following functions:

5 / (7 x - 3)

2 t / (3 t^2 - 5)

Identify the values of a, b and c for which the function 5 / (7 x - 3) is of the form c / (ax + b), then use the formula to write down the integral as given by the table.

 

For the second function you will need to do a substitution to put the integral into the required form.  Using the substitution u = t^2, express c / (ax + b) dx in terms of u and du, apply the formula, and then convert your expression back to the variable t.

 

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Your solution:

 5 / (7 x - 3)

A = 7

B = -3

C = 5

(5/7) * ln|7x -3| + C

2 t / (3 t^2 - 5)

U = t^2

Du = 2t

Int(Du/(u-5)) = ln(t^2 - 5)

confidence rating #$&*:

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Self-critique Rating:OK

 

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Question:  `q002.  Use the formula

 

integral ( sin^2(x) dx ) = 1/2 ( x - sin(2x) / 2) + C

 

to integrate e^(2 t) sin^2 ( e^(2 t) dt).

 

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Your solution:

 U = e^2t

Du = 2e^2t

Int(sin^2(u)) = 1/2 ( e^2t - sin(2e^2t) / 2) + C

confidence rating #$&*:

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Self-critique Rating:

 

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Question:  `q003.  Use the formula

 

integral ( cos(ax) e^(bx) dx ) = e^(b x) / (a^2 + b^2) (a sin(ax) + b cos(ax) ) + C

 

to find

 

integral ( cos ( 5 x) e^(-2 x) dx ).

 

 

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Your solution:

 

confidence rating #$&*:

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Self-critique Rating:

 

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Question:  `q004.  Solve the equation

 

A / (x + 1) + B / (x - 2) = (3x + 2) / ( (x + 1) ( x - 2) ).

 

So solve, express both of the fractions on the left-hand side with the common denominator (x + 1) ( x - 2), express the sum as a single fraction with this denominator. 

 

Set the numerators of both sides equal.  If the equation is to be true, it will be true for all values of x, so the coefficient of x on the left will match the coefficient of x on the right, and the constant quantity on the left will match the constant quantity on the right.  This will give you two simultaneous equations in A and B, which you will need to solve.

 

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Your solution:

 

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Self-critique Rating:

 

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Question:  `q005.  Integrate the expression

 

3 / (x + 3) + 5 / (x - 4).

 

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Your solution:

 

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Self-critique Rating:

 

 

Questions from text assignment:

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Question:  Section 7.3 Problem 3

 

problem 7.3.3 (previously 7.3.15)  x^4 e^(3x)  ****   what it is your antiderivative?

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                                                  20:35:16

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

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                                                  20:35:18

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Self-critique (if necessary): OK

 

 

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Self-critique Rating:

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Question:   Which formula from the table did you use?

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Your solution:

 

 

confidence rating #$&*: OK

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Given Solution:

You should have used formula 14, with a = 3 and p(x) = x^4. 

 

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Self-critique (if necessary):

 

 

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Self-critique Rating:

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Question:   Question 7.3 Problem 7

 

problem 7.3.7 (previously 7.3.33  1 / [ 1 + (z+2)^2 ) ])  ****   What is your integral?  ****   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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Your solution:

Rule 26

 Int( 1/(x-a)(x-b)) ?

 

 

confidence rating #$&*: 1

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Given Solution:

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 using techniques of integration you should already be familiar with: 

 

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).

 

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Self-critique (if necessary):

 Needed to rearrange the integrand to use another formula.

 

 

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Self-critique Rating:3

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Question:  Problem 7.4 Problem 1

   7.4.1 (previously 7.4.6).  Integrate 2y / ( y^3 - y^2 + y - 1)

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Your solution:

 2y/(y^2 + 1)(y-1)

 (Ay+B/Y^2 + 1) + (C/y-1)

 Ay^2 - Ay + By - B + Cy^2 + C = 2y

 (A + C)y^2 + (-A + B)y - B + C = 2y

 A + C = 0

 -A + B = 2

 -B + C =0

 A = -1; B = 1; C = 1

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

 U = y^2

 Du = 2ydy

 -du/2 = -ydy

 Int(1 / (u + 1))

 -(1/2)ln(y^2+1)

 Int((1/y^2 + 1))

 Arctan(y)

 Int((1/(y-1))

 Ln(y-1)

 Combine:

 -(1/2)ln(y^2+1) + Arctan(y) + Ln(y-1) + C

confidence rating #$&*: 3

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Given Solution:

 

Let's integrate just y / (y^3 - y^2 + y - 1), then double the result.

 

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 so that 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 ) / (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

 

-1/4 ln(y^2 + 1) + 1/2 arcTan(y) + 1/2 ln | y - 1 |

Doubling the result to get the integral of the given function we have

 

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

 

where c now stands for an arbitrary integration constant.

 

DER

 

 

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Self-critique (if necessary): OK, solution integrand is not the same as the original integrand.

 

 

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Self-critique Rating:

 

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Question:  Section 7.4 Problem 7

 

7.4.12 (previously 7.4.29 (4th edition)).  Integrate (z-1)/`sqrt(2z-z^2)  ****   What  did you get for your integral?

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Your solution:

 U = 2z - z^2

 Du = 2 - 2z

 Du/-2 = -1 + z || z -1dz

 1/u^(-1/2)

-(2z -z^2)^(1/2)/2

 

 

confidence rating #$&*:3

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Given Solution:

 

 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).

 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).

DER

 

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Self-critique (if necessary): I made some algebraic mistakes. I had the exponent as negative for u while it was in the denominator.

 

 

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Self-critique Rating:3

 

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Question:  Section 7.4 Problem 9

 

7.4.9 (previously 7.4.36) partial fractions for 1 / (x (L-x))

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Your solution:

 (1/x) + (1/L-x)

 (a/x) + (b/L-x)

 aL - ax + bx = 1

 aL = 1

 -a + b = 0

 A = 1/L

 1/L + b = 0

 b = -1/L

 int[(1/Lx) - (1/L^2 - Lx)]

 

 

confidence rating #$&*: 1

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Given Solution:

a / x + b / (L-x) = [ a (L-x) + bx ] / [ x(L-x)] = [ a L + (b-a)x ] / [ x(L-x)].

This is equal to 1 / [ x(L-x) ].

So a L = 1 and (b-a) = 0.

Thus a = 1 / L, and since b-a=0, b = 1/L.

The original function is therefore 1 / x + b / (L-x) = 1 / L [ 1 / x + 1 / (L-x) ].

Integrating we get 1 / L ( ln(x) - ln(L-x) ) = 1 / L ln(x / (L-x) ). **

 

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Self-critique (if necessary): Don’t understand your process of integration at the end with two variables involved

@&

The only variable is x.

Antiderivative of 1/x is ln | x |.

Antiderivative of 1 / (L - x) is - ln( |L - x| ), which by laws of logs is equal to ln ( 1 / | L - x | ) If you don't see this integral use the substitution u = L - x.

*@

 

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Self-critique Rating: 3

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Question:  Section 7.4 Problem 6

 

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

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Your solution:

Factored out:

 (y+2)/(2y+1)(y+1)

 A/(2y+1) + B/(y+1)

Numerators combined:

 Ay + A + 2By + B = y + 2

Solving for A and B

 (A + 2B) + A + B

A + 2B = 1

A + B = 2

 A = 3

 B = -1

 Int((3/2)(1/y+1)-(1/y+1)) //logs

U = 2y + 1

Du = 2dy

Du/2 = dy

(3/2)int(1/u) = (3/2)ln(2y+1)

 (3/2)ln(2y+1) - ln(y+1)

 

  

 

 

confidence rating #$&*:3

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Given Solution:

(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.

 

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Self-critique (if necessary): The terms in the denominator are being added; does it still work with formula 27 in the book? I don’t see where I went wrong with my work.

 

 

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Self-critique Rating:3

@&

The correct result is

1/2 ln ((2y + 1)^3/(y + 1)^2) ,

which can also be expressed as

3/2 ln(2y+1) - ln(y+1)

This is also what you get if you substitute the values of a, b, c and d in the given solution.

So your solution is both correct and consistent with the given solution.

*@

 

STUDENT COMMENTS:

 

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?

 

INSTRUCTOR RESPONSE:

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.

 

CONTINUED STUDENT COMMENT:

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).

 

INSTRUCTOR RESPONSE

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.

 

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Question:  `q006.  Integrate (t + 3) / (t^2 - 5 t + 6).

 

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Your solution:

 (t+3)/(t-2)(t-3)

a/(t-2) + b/(t-3)

at - 3a + bt -2b = t + 3

(a + b)t -3a -2b = t + 3

A+ B = 1

-3a -2b = 3

A=-5

B=6

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

-5ln(t-2)+6ln(t-3)

confidence rating #$&*:

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Self-critique (if necessary):

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#*&!

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