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Q: Help with calcuclus 1 problems ($100) ( Answered 3 out of 5 stars,   1 Comment )
Subject: Help with calcuclus 1 problems ($100)
Category: Science > Math
Asked by: alwin86-ga
List Price: $100.00
Posted: 11 May 2006 09:37 PDT
Expires: 10 Jun 2006 09:37 PDT
Question ID: 727714
I need somebody to solve these 5 calculus 1 problems so I can
understand them. It's a practice sheet my teacher gave me. Im willing
to pay $100 to anyone who can solve them. I need it before Friday 10

Here are two links with the problems in two different images.
Thank you!!
Subject: Re: Help with calculus 1 problems
Answered By: leapinglizard-ga on 11 May 2006 12:49 PDT
Rated:3 out of 5 stars
Dear alwin86,

Detailed solutions follow. Remember, these are for study purposes
only. You must not attempt to pass them off as your own work.


The distance between a point (x, y) and the point (4, 0) is the square
root of

  (x - 4)^2 + y^2 .

Substituting y = f(x) = sqrt(x) in this expression, we see that

  g(x)  =  x^2 - 8x + 16 + x

        =  x^2 - 7x + 16

is the square of the distance.

To minimize the distance, we find the first derivative

  g'(x)  =  2x -  7

which tells us that when

  2x - 7  =  0

       x  =  3.5

the parabola g(x) reaches its lowest point. Thus, the point

  (3.5, sqrt(3.5))

is the point on f(x) closest to (4, 0).


The average value of f(x) = x^3 over [0, 2] is the same as its average
height, which is the area under the curve divided by 2, the length of
the interval.

The first integral of f(x) is

  x^4 / 4

which tells us that the area under f(x) from 0 to 2 is

  2^4 / 4  =  16 / 4

           =  4 .

Dividing this by 2, we obtain

  4 / 2  =  2

as the average value of f(x) over the interval [0, 2]. The function
assumes this value when

  x^3  =  2

    x  =  2^(1/3)

       =  1.26

with two decimal digits of precision, so

    (1.26, 2)

is approximately where the curve passes through its average value over
the interval [0, 2].


The graph of f(x) lies below the x-axis in the interval [0, 1] and over
the x-axis in [1, 2]. We must therefore calculate each area separately
before summing.

The first integral of f(x) is

  x^3 / 3 - x .

Evaluated at x = 1, this gives us

  1^3 / 3 - 1  =  1/3 - 1

               =  -2/3

so the area between the curve and the x-axis is 2/3 in the interval
[0, 1].

Evaluating the integral at x = 2, we have

  2^3 / 3 - 2  =  8/3 - 2

               =  2/3

and subtracting the earlier obtained -2/3 at x = 1, we obtain

  2/3 - (-2/3)  =  2/3 + 2/3

                =  4/3

as the area under the curve in the interval [1, 2].

To find the total area between the curve and the x-axis in the interval
[0, 2], we sum the areas to obtain

  2/3 + 4/3  =  2 .


According to the second fundamental theorem of calculus --

Mathworld: Second Fundamental Theorem of Calculus

-- the first derivative of F(x) is simply the function under the integral
evaluated at t = x.

  F'(x)  =  1/x^2


The first integral of the function y = 1 - x^4 is

  x - x^5 / 5 .

In the interval [0, 1], the area under the curve is therefore

  1 - 1^5 / 5  =  1 - 1/5

               =  4/5 .

Because the curve is symmetrical, the area covered in [-1, 0] will be
the same.

Thus, the total area covered in the interval [-1, 1] is

  4/5 + 4/5  =  8/5 .

If you have any concerns about the accuracy of these solutions, please
advise me through a Clarification Request and allow me the opportunity
to fully meet your needs before you rate this answer.


Search strategy:  

second fundamental theorem calculus

Request for Answer Clarification by alwin86-ga on 11 May 2006 13:26 PDT
This loosk very good, but  I would like to have a few more details.
For example in question number one:

The distance between a point (x, y) and the point (4, 0) is the square
root of

How did you get:
  (x - 4)^2 + y^2 .

Did you  do this on a paper, would it be possible to see the paper you
wrote on so maybe I can get a better idea. There are more things Im
not sure how to get. So i dont know if you could give a few more
details.  This still is a very good job and after a few more questions
im sure I'll accept this answer..

Clarification of Answer by leapinglizard-ga on 11 May 2006 13:44 PDT
The distance between points (x0, y0) and (x1, y1) is given by
sqrt((x0-y0)^2 + (x1-y1)^2). If y1 is 0, then we can simplify it to
sqrt((x0-y0)^2 + x1^2). See here for an in-depth explanation of the

Mathworld: Point-Point Distance--2-Dimensional

I cannot give you any work on paper, but we can continue to
communicate through the Clarification Request mechanism. Let me know
if you need any further explanations.


Request for Answer Clarification by alwin86-ga on 11 May 2006 15:43 PDT
Hello, Thanks to give the extra explanation, I am going to keep doing
the rest of the problems if I have a question I hope you can still
answering me. Besides that I have other question. Would you do other
problem for an extra 30 dollars. Please let me know.

Clarification of Answer by leapinglizard-ga on 11 May 2006 16:10 PDT
I remain at your disposal. If you post a further problem, I'll take a look at it.


Request for Answer Clarification by alwin86-ga on 11 May 2006 16:18 PDT
Hello, I have a question. in number 3 its says to find the area of the
region using the limit method. did you used it? please let me know.

and i'm going to post the other problem. I hope you get it first.

Request for Answer Clarification by alwin86-ga on 11 May 2006 16:35 PDT
I put the new problem up. please I would like you to help me with this
problem also. Thank You..

Clarification of Answer by leapinglizard-ga on 11 May 2006 18:00 PDT
In problem 3, our use of limits is implied in taking the first
integral of f(x). We do not apply the limit method directly, as we
would in taking the first derivative of a function. An alternative to
integrating f(x) would be to analyze the graph geometrically, for
instance by considering the graph of g(x) = f(x) + 1 and dividing the
area under g(x) into three quadrants, but that's not what I did here.

alwin86-ga rated this answer:3 out of 5 stars
Need to do ALL problems as the questions ask

Subject: Re: Help with calcuclus 1 problems ($100)
From: research_help-ga on 11 May 2006 13:34 PDT
Does it seem strange to anyone else besides me that someone is paying
$100 for help studying with a few calculus problems with a deadline of
"Friday 10 am"?  Doesn't this raise some really large red flags waving
in a strong wind on the tallest flagpole?

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