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Q: Help with calcuclus 1 problems (\$100) ( Answered ,   1 Comment )
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 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 am. Here are two links with the problems in two different images. http://www.imagenspace.com/image/11514.jpeg http://www.imagenspace.com/image/11516.jpeg Thank you!!```
 ```Dear alwin86, Detailed solutions follow. Remember, these are for study purposes only. You must not attempt to pass them off as your own work. 1. 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). 2. 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]. 3. 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 . 4. According to the second fundamental theorem of calculus -- Mathworld: Second Fundamental Theorem of Calculus http://mathworld.wolfram.com/SecondFundamentalTheoremofCalculus.html -- the first derivative of F(x) is simply the function under the integral evaluated at t = x. F'(x) = 1/x^2 5. 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. Regards, leapinglizard Search strategy: second fundamental theorem calculus ://www.google.com/search?q=second+fundamental+theorem+calculus&start=0&ie=utf-8&oe=utf-8&client=firefox-a&rls=org.mozilla:en-US:official``` 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 formula. Mathworld: Point-Point Distance--2-Dimensional http://mathworld.wolfram.com/Point-PointDistance2-Dimensional.html 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. leapinglizard``` 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. Thanks?``` 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. leapinglizard``` 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. thanksss 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. leapinglizard```
 alwin86-ga rated this answer: `Need to do ALL problems as the questions ask`
 ```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?```