Assignment 1

course Mth 174

RܷYՊTassignment #001

د`SB

Physics II

06-01-2009

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16:40:24

What was your value for the integral of f'?

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RESPONSE -->

The value I got fot the integral of f was 8. I know this is not the right answer and I'm not sure where I made a mistake. This is how I set up the problem:

f(3)=0

f(2)= integral f'(x)dx - f(3) = .5 * 1 * 1 - 0 = .5

f(0)= integral f'(x)dx - f(2) = .5 * 1 *1 - .5 = 0

f(3)=0

f(4)= f(3) + integral f'(x)dx= 0 + .5 * 1 * 2 = 1

f(5) = f(4) + integral f'(x)dx= 1 + .5 * 1 * 2= 2

f(6) = f(5) - integral f'(x)dx= 2 - .5 * 1 * -2 = 3

f(7) = f(6) - integral f'(x)dx= 3 - .5 * 1 * 1 = 8

I used the antiderivative graph I drew from the graph given.

the change in f from x=0 to x=2 is 2 (area beneath line segment from x=0 to x=1), then from x=2 to x=3 is -1, then from x=3 to x=4 is +2, then from x=4 to x=6 is -4, then from x=6 to x=7 is +1.

If f(3) = 0 then f(4) = 0 + 2 = 2, f(6) = 2 - 4 = -2 and f(7) = f(6) + 1 = -1.

Working back from x=3, f(2) = 0 - (-1) = 1 and f(0) = 1 - 2 = -1.

The integral is the sum of the changes in f ' which is 2 - 1 + 2 - 4 + 1 = 0.

Alternatively since f(0) = -1 and f(7) = -1 the integral is the difference f(7) - f(0) = -1 - (-1) = 0.

Let me know if you disagree with or don't understand any of this and I will explain further. Let me know specifically what you do and don't understand.

Alternative solution method:

Two principles will allow you to solve this problem:

1. The definite integral of f' between two points gives you the change in f between those points.

2. The definite integral of f' between two points is represented by the 'signed area' beneath the graph of f' between the two points, provided 'signed area' is understood as being positive when the graph is above the x axis and negative when the graph is below.

We apply these two principles to determine the change in f over each of the given intervals.

To apply these principles you would answer the following questions:

What is the area beneath the graph of f' between x = 0 and x = 2?

What is the area beneath the graph of f' between x = 3 and x = 4?

What is the area beneath the graph of f' between x = 4 and x = 6?

What is the area beneath the graph of f' between x = 6 and x = 7?

What is the change in the value of f between x = 3 and x = 4? Since f(3) = 0, what therefore is the value of f at x = 4?

Now that you know the value of f at x = 4, what is the change in f between x = 4 and x = 6, and what therefore is the value of f at x = 6?

Using similar reasoning, what is the value of f at x = 7?

Then using similar reasoning, see if you can determine the value of f at x = 2 and at x = 0.**

STUDENT QUESTION: I did not understand how to obtain the value of f(0), but I found that f(7)

was 10 by adding all the integrals together

INSTRUCTOR RESPONSE:

The total area is indeed 10, so you're very nearly correct; however the integral is like a 'signed' area--areas beneath the x axis make negative contributions to the integral--and you added the 'absolute' areas

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16:40:53

What was the value of f(0), and of f(7)?

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RESPONSE -->

The value I got fot the integral of f was 8. I know this is not the right answer and I'm not sure where I made a mistake. This is how I set up the problem:

f(3)=0

f(2)= integral f'(x)dx - f(3) = .5 * 1 * 1 - 0 = .5

f(0)= integral f'(x)dx - f(2) = .5 * 1 *1 - .5 = 0

f(3)=0

f(4)= f(3) + integral f'(x)dx= 0 + .5 * 1 * 2 = 1

f(5) = f(4) + integral f'(x)dx= 1 + .5 * 1 * 2= 2

f(6) = f(5) - integral f'(x)dx= 2 - .5 * 1 * -2 = 3

f(7) = f(6) - integral f'(x)dx= 3 - .5 * 1 * 1 = 8

I used the antiderivative graph I drew from the graph given.

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16:43:12

Describe your graph of f(x), indicating where it is increasing and decreasing and where it is concave up, where it is straight and where it is concave down.

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RESPONSE -->

There are no concaves, but each line is linear because the lines in the f' graph are constant.

Increasing from (0, -1) to (2,1) then decreasing to (3,0), then increasing to (4,2), then decreasing to (6,-2), then increasing to (7,-1)

** The graph of f(x) is

increasing, with slope 1, on the interval (0,2), since f'(x) = 1 on that interval,

decreasing, with slope -1, on the interval (2,3), where f'(x) = -1,

increasing, with slope +2, on the interval (3,4), where f'(x) = +2,

decreasing, with slope -2, on the interval (4,6), where f'(x) = -2, and

increasing, with slope +1, on the interval (6,7), where f'(x) = +1.

The concavity on every interval is zero, since the slope is constant on every interval.

Since f(3) = 0, f(4) = 2 (slope 2 from x=3 to x=4), f(6) = -2 (slope -2 from x = 4 to x = 6), f(7) = -1 (slope +1 from x=6 to x=7).

Also, since slope is -1 from x=2 to x=3, f(2) = +1; and similar reasoning shows that f(0) = -1. **

** The definite integral of f'(x) from x=0 to x=7 is therefore f(7) - f(0) = -1 - (-1) = 0. **

** Basic principles:

1. The slope of the graph of f(x) is f'(x). So the slope of your f graph will be the value taken by your f' graph.

2. Note that if the slope of the f graph is constant for an interval that means that the graph is a straight line on the interval.

Using these principles answer the following questions:

What is the slope of the f graph between x = 0 and x = 2?

What is the slope of the f graph between x = 3 and x = 4?

What is the slope of the f graph between x = 4 and x = 6?

What is the slope of the f graph between x = 6 and x = 7?

Given that f(3) = 0 and using the value of the slope of the f graph between x = 3 and x = 4, describe the f graph between these two points.

Using similar information describe the graph for each of the other given intervals.

Also answer the following:

What would have to be true of the f' graph for the f graph to be concave up? Same question for concave down. **

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16:43:34

Was the graph of f(x) continuous?

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RESPONSE -->

This graph was continuous.

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16:45:12

How can the graph of f(x) be continuous when the graph of f'(x) is not continuous?

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RESPONSE -->

This occurs because the graph f '(x) is only a representation of the slopes of the f(x). Places where the f '(x) graph is not continuous shows that it is linear with no concaves.

** f ' is the slope of the f graph; f ' has 'jumps', which imply sudden changes in the slope of the f graph, causing the graph of f to have a jagged shape as opposed to a smooth shape. However this does not cause the graph of f itself to have discontinuous 'jumps'. **

** f ' determines the slope of f; the slope of f can change instantaneously without causing a 'jump' in the values of f. Continuity is, roughly speaking, a lack of 'jumps' in a graph. **

** Basically, if f ' is finite and does exceed some fixed bound over a small interval about x = a, then the change `dx in x has to be small. More specifically:

f(x) is continuous at x = a if the limit of f(x) as x -> a is equal to f(a).

If f (x) is bounded in some vicinity of x = a, then this condition must be satisfied. Specifically if for | x a | < epsilon we have | f | < L, it follows that on this same interval | f(x) f(a) | < epsilon * L. So as x -> a, f(x) -> f(a) and the function f is continuous at a.

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16:45:56

What does the graph of f(x) look like over an interval where f'(x) is constant?

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RESPONSE -->

When f '(x) is constant, the raph of f(x) has concaves.

** If f ' is constant then the slope of the f(x) graph is constant, so the graph of f(x) must be linear **

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16:51:52

What were the areas corresponding to each of the four intervals over which f '(x) was constant? What did each interval contribute to the integral of f ' (x)?

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RESPONSE -->

These are the areas I found and each interval is either added to or subtracted from the interval depending on if it is below or above the x-axis.

f(3)=0

f(2)= integral f'(x)dx - f(3) = .5 * 1 * 1 - 0 = .5

f(0)= integral f'(x)dx - f(2) = .5 * 1 *1 - .5 = 0

f(4)= f(3) + integral f'(x)dx= 0 + .5 * 1 * 2 = 1

f(5) = f(4) + integral f'(x)dx= 1 + .5 * 1 * 2= 2

f(6) = f(5) - integral f'(x)dx= 2 - .5 * 1 * -2 = 3

f(7) = f(6) - integral f'(x)dx= 3 - .5 * 1 * 1 = 8

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16:53:59

the change in f from x=0 to x=2 is 2 (area beneath line segment from x=0 to x=1), then from x=2 to x=3 is -1, then from x=3 to x=4 is +2, then from x=4 to x=6 is -4, then from x=6 to x=7 is +1.

If f(3) = 0 then f(4) = 0 + 2 = 2, f(6) = 2 - 4 = -2 and f(7) = f(6) + 1 = -1.

Working back from x=3, f(2) = 0 - (-1) = 1 and f(0) = 1 - 2 = -1.

The integral is the sum of the changes in f ' which is 2 - 1 + 2 - 4 + 1 = 0.

Alternatively since f(0) = -1 and f(7) = -1 the integral is the difference f(7) - f(0) = -1 - (-1) = 0.

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RESPONSE -->

I used the graph I made instead of the one shown in the book. I understand how to add up the areas.

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16:57:26

When was the quantity of water greatest and when least? Describe in terms of the behavior of the two curves.

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RESPONSE -->

Greatest around April and May because here the inflow is greater than the outflow.

Least in October because here inflow i decreasing greatly while outflow is at its greatest.

If outflow is greater than inflow, there is less water and if inflow is greater than outflow there is more water.

** Between any two dates the corresponding outflow is represented by the area under the outflow curve, and the inflow by the area under the inflow curve.

When inflow is greater than outflow the quantity of water in the reservoir will be increasing and when the outflow is greater than the inflow quantity of water will be decreasing.

We see that the quantity is therefore decreasing from January 93 through sometime in late February, increasing from late February through the beginning of July, then again decreasing through the end of the year.

The reservoir will reach a relative maximum at the beginning of July, when the outflow rate overtakes the inflow rate.

The amount of water lost between January and late February is represented by the difference between the area under the outflow curve and the area under the inflow curve. This area corresponds to the area between the two graphs. The amount of water gained between late February and early July is similarly represented by the area between the two curves. The latter area is clearly greater than the former, so the quantity of water in the reservoir will be greater in early July than on Jan 1.

The loss between July 93 and Jan 94, represented by the area between the two graphs over this period, is greater than the gain between late February and early July, so the minimum quantity will occur in Jan 94.

The rate at which the water quantity is changing is the difference between outflow and inflow rates. Specifically the net rate at which water quantity is changing is

net rate = inflow rate - outflow rate.

This quantity is represented by the difference between the vertical coordinate so the graphs, and is maximized around late April or early May, when the inflow rate most greatly exceeds the outflow rate. The net rate is minimized around early October, when the outflow rate most greatly exceeds the inflow rate. At this point the rate of decrease will be maximized. **

SOME ADDITIONAL QUESTIONS TO CONSIDER:

** When inflow is > outflow the amount of water in the reservoir will be increasing. If outflow is < inflow the amount of water will be decreasing.

Over what time interval(s) is the amount of water increasing?

Over time interval(s) is the amount of water decreasing? **

** What aspect of which graph gives you the rate at which water is flowing into the reservoir?

What aspect of which graph gives you the rate at which water is flowing out of the reservoir?

What has to be true of the two graphs in order for the amount of water in the reservoir to the increasing at an increasing rate?

What has to be true of the two graphs in order for the amount of water in the reservoir to the increasing at a decreasing rate?

What has to be true of the two graphs in order for the amount of water in the reservoir to the decreasing at an increasing rate?

What has to be true of the two graphs in order for the amount of water in the reservoir to the decreasing at a decreasing rate? **

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16:58:33

When was the quantity of water increasing fastest, and when most slowly? Describe in terms of the behavior of the two curves.

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RESPONSE -->

Increasing the fastest from Jan to April because inflow is increasing more rapidly than outflow.

Most slowly from July to Oct because inflow is decreasing and outflow is increasing.

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17:01:08

Between any two dates the corresponding outflow is represented by the area under the outflow curve, and the inflow by the area under the inflow curve.

When inflow is greater than outflow the quantity of water in the reservoir will be increasing and when the outflow is greater than the inflow quantity of water will be decreasing.

We see that the quantity is therefore decreasing from January 93 through sometime in late February, increasing from late February through the beginning of July, then again decreasing through the end of the year.

The reservoir will reach a relative maximum at the beginning of July, when the outflow rate overtakes the inflow rate.

The amount of water lost between January and late February is represented by the difference between the area under the outflow curve and the area under the inflow curve. This area corresponds to the area between the two graphs. The amount of water gained between late February and early July is similarly represented by the area between the two curves. The latter area is clearly greater than the former, so the quantity of water in the reservoir will be greater in early July than on Jan 1.

The loss between July 93 and Jan 94, represented by the area between the two graphs over this period, is greater than the gain between late February and early July, so the minimum quantity will occur in Jan 94.

The rate at which the water quantity is changing is the difference between outflow and inflow rates. Specifically the net rate at which water quantity is changing is

net rate = inflow rate - outflow rate.

This quantity is represented by the difference between the vertical coordinate so the graphs, and is maximized around late April or early May, when the inflow rate most greatly exceeds the outflow rate. The net rate is minimized around early October, when the outflow rate most greatly exceeds the inflow rate. At this point the rate of decrease will be maximized.

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RESPONSE -->

I understand this problem.

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17:04:21

antiderivative of f(x) = x^2, F(0) = 0

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RESPONSE -->

The problem in my book was different. It was

f(x)= sqrt' x

integral of sqrt'x dx =

(x^ (3/2)) / 3/2 + c

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17:04:45

What was your antiderivative? How many possible answers are there to this question?

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RESPONSE -->

The problem in my book was different. It was

f(x)= sqrt' x

integral of sqrt'x dx =

(x^ (3/2)) / 3/2 + c

There are infinant possible answers becase c can equal any constant number.

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17:05:26

What in general do you get for an antiderivative of f(x) = x^2?

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RESPONSE -->

f(x) = x^2

integral of x^2 dx=

x^3/3

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17:06:37

An antiderivative of x^2 is x^3/3.

The general antiderivative of x^2 is F(x) = x^3/3 + c, where c can be anything. There are infinitely many possible specific antiderivative.

However only one of them satisfied F(0) = 0. We have

F(0) = 0 so 0^3/3 + c = 0, or just c = 0.

The antiderivative that satisfies the conditions of this problem is therefore F(x) = x^3/3 + 0, or just F(x) = x^3/3.

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RESPONSE -->

When F(0) = 0 than c must = 0.

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17:09:55

Query Section 6.2 #55 (3d edition #56) indef integral of t `sqrt(t) + 1 / (t `sqrt(t))

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RESPONSE -->

This one was a bit confusing for me. This is how I set it up:

integral of t * t^1/2 + integral of -(t * t^1/2)=

t^2/2 *t^(3/2) / 3/2 - t^2/2 * t^(3/2) / 3/2

I wasn't sure what to do after this.

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17:10:00

What did you get for the indefinite integral?

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RESPONSE -->

This one was a bit confusing for me. This is how I set it up:

integral of t * t^1/2 + integral of -(t * t^1/2)=

t^2/2 *t^(3/2) / 3/2 - t^2/2 * t^(3/2) / 3/2

I wasn't sure what to do after this.

The function can be written t^(3/2) + t^(-3/2). Both are power functions of the form t^n. Antiderivative is

2/5 * t^(5/2) - 2 t^(-1/2) + c or

2/5 t^(5/2) - 2 / `sqrt(t) + c. **

(2/5) t^(5/2) + ln(t^3/2)

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17:30:07

What is an antiderivative of t `sqrt(t)?

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RESPONSE -->

integral of t `sqrt(t) = t^(3/2) / (3/2)

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17:31:12

What is an antiderivative of 1/(t `sqrt(t))?

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RESPONSE -->

antiderivative of 1/(t `sqrt(t)) =

- t sqrt t^ (3/2) / (3/2)

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17:32:14

What power of t is t `sqrt(t)?

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RESPONSE -->

t^ 1/2 = sqrt (t)

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17:32:44

What power of t is 1/(t `sqrt(t))?

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RESPONSE -->

t = sqrt t^-1/2

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17:40:27

The function can be written t^(3/2) + t^(-3/2). Both are power functions of the form t^n. Antiderivative is

2/5 * t^(5/2) - 2 t^(-1/2) + c or

2/5 t^(5/2) - 2 / `sqrt(t) + c.

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RESPONSE -->

I used 1/2 instead of 3/2.

sqrt(t) = t^ 3/2, not t^1/2

I am still confused on what to do from here:

t^2/2 * 2/5 * t^5/2 - t^2/2 * 2t^-1/2

what happened to the first t^2/2?

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17:45:58

Query Section 6.2 #68 (3d edition #69) def integral of sin(t) + cos(t), 0 to `pi/4

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RESPONSE -->

def integral of sin(t) + cos(t), 0 to `pi/4

def integral of sin(t) dt, 1 tp pi/4 + def integral of cos(t) dt, 0 to pi/4=

-cos(t) + sin(t), 0 to pi/4

-cos(0) + sin(0)= -2

-cos(pi/4) + sin(pi/4) = 0

-2 - 0 = -2

** An antiderivative is -cos(t) + sin(t), as you can see by taking the derivative.

Evaluating this expression at `pi/4 gives -`sqrt(2)/2 + `sqrt(2)/2 = 0. Evaluating at 0 gives -1 + 0 or -1. The antiderivative is therefore 0 - (-1) = 1. **

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17:46:37

What did you get for your exact value of the definite integral?

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RESPONSE -->

def integral of sin(t) + cos(t), 0 to `pi/4

def integral of sin(t) dt, 1 tp pi/4 + def integral of cos(t) dt, 0 to pi/4=

-cos(t) + sin(t), 0 to pi/4

-cos(0) + sin(0)= -2

-cos(pi/4) + sin(pi/4) = 0

-2 - 0 = -2

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17:46:47

What was your numerical value?

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RESPONSE -->

-2 - 0 = -2

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17:47:15

What is an antiderivative of sin(t) + cos(t)?

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RESPONSE -->

-cos(t) + sin(t) + c

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17:47:40

Why doesn't it matter which antiderivative you use?

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RESPONSE -->

They both equal each other no matter what constant you use.

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17:48:58

An antiderivative is -cos(t) + sin(t), as you can see by taking the derivative.

Evaluating this expression at `pi/4 gives -`sqrt(2)/2 + `sqrt(2)/2 = 0. Evaluating at 0 gives -1 + 0 or -1. The antiderivative is therefore 0 - (-1) = 1.

The general antiderivative is -cos(t) + sin(t) + c, where c can be any number. You would probably use c = 0, but you could use any fixed value of c.

Since c is the same at both limits of the integral, it subtracts out and has no effect on the value of the definite integral.

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RESPONSE -->

I miscalculated sin(0). I reworked this part of the problem and got 0. I understood the rest of the problem.

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17:50:59

Query Section 6.2 #82 (#81 3d edition) v(x) = 6/x^2 has average value 1 on [1,c}; find c

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RESPONSE -->

v(x) = 6/x^2 +c on [1,c]

6/(1)^2 +c = 1

6 + c = 1

c= -5

An antiderivative of 6 / x^2 is F(x) = -6 / x.

The definite integral is equal to the product of the average value and the length of the interval. In this case average value is 1 and the interval from x = 1 to x = c has length c 1. So the definite integral must be 1 * ( c 1).

Evaluating between 1 and c and using the above fact that the result must be 1 we get

F(c) - F(1) = -6/c- (-6/1) = c - 1 so that

-6/c+6=c - 1. We solve for c, first getting all terms on one side:

c 7 + 6/c = 0. Multiplying both sides by c to get

c^2 7 c + 6 = 0. Either be factoring or the quadratic formula we get

c = 6 or c = 1.

If c = 1 the interval has length 0 and the definite integral is not defined. This leaves the solution

c= 6.

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17:51:04

What is your value of c?

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RESPONSE -->

v(x) = 6/x^2 +c on [1,c]

6/(1)^2 +c = 1

6 + c = 1

c= -5

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18:09:55

In symbols, what did you get for the integral of 6 / x^2 over the interval [1, c]?

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RESPONSE -->

I didn't integrate before but now I think I understand how to do it.

integral of 6 / x^2

integral of 6 * x^3/3 + 1

6 * 1^3/3 +1 + c= -3

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18:15:22

An antiderivative of 6 / x^2 is F(x) = -6 / x.

Evaluating between 1 and c and noting that the result must be 1 we get

F(c) - F(1) = -6/c- (-6/1) = 1 so that

-6/c+6=1. We solve for c:

-6/c=1-6

6/c=-5

-6=-5c

c=6/5.

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RESPONSE -->

I multiplied instead of divided 6.

I should have gotten g * -x^-1 and simplified to -6/x.

Then I should have used F(c) - F(1) = -6/c- (-6/1) = 1 to find c.

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18:45:02

Extra Problem (formerly from Section 6.2 #44): What is the indefinite integral of e^(5+x) + e^(5x)

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RESPONSE -->

indefinite integral of e^(5+x) + e^(5x)=

indefinite integral of e^(5 + x) + integral of e^(5x)=

I used substitution here

u = 5 + x

du = 1

w=5x

dw=5

integral of e^u du + integral e^w * dw/5 =

integral of e^u du + 1/5 integral e^w dw=

e^(5 + x) + 1/5 * e^(5x) + c

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18:45:34

The derivative of e^(5+x) is, by the Chain Rule, (5+x)' * e^(5+x) = 1 * e^(5 + x) = e^(5 + x) so this function is its own antiderivative.

The derivative of e^(5x) is (5x) ' * e^(5x) = 5 * e^(5x). So to get an antiderivative of e^(5x) you would have to use 1/5 e^(5x), whose derivative is e^(5x). **

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RESPONSE -->

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