Calculating the Area of a Strip Using a Riemann Sum

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To calculate the area of the strip between the curves defined by the equations 6x + y = 12 and y = x^2 - 4, the limits of integration are established from x = 0 to x = 2. The upper curve is y = 12 - 6x, which is always above the lower curve y = x^2 - 4 within this interval. The area can be determined by integrating the difference between the upper and lower functions over the specified bounds, resulting in the integral of (16 - 6x - x^2) from 0 to 2. A Riemann sum can also be used to approximate this area by partitioning the interval and calculating the sum of rectangle areas. The discussion emphasizes the importance of identifying which curve is above the other to set up the integral correctly.
Mgeorges
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Homework Statement




Write a Riemann sum and then a definite integral representing the area of the region, using the strip shown in the figure below where the upper line is defined by 6x + y = 12 and the other line is defined by y=x^2-4. The figure, which I can't get on here, is just the area bounded between those two equations. I do not need the Riemann sum, I just need to find: (a) What is the approximate area of the strip with respect to x (the strip is horizontal)? I found the limits of integration which is from [0,2]

Homework Equations





The Attempt at a Solution


I have no idea how to find the area of the strip, and after that I can figure out the integral with no problem.
 
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If you can find the integral with no problems, how come you need to find out the area of the strip first? And those two curves intersect at x=2 but not x=0. Did the figure you didn't show only tell you to consider x>=0?
 


I have to find the area of the strip, and from there I can get the integral..The method were using is volumes by slicing, and the [0,2] were the bounds for the integral. I do not know how to set up this definite integral anyway.
 


To get a Riemann sum you create a partition of [0,2] and then calculate an upper sum or a lower sum to approximate the area of the region. It's just a sum of rectangle areas. If you want to set up an integral you have to figure out which curve is above the other in the x interval [0,2], subtract the lower value from the upper value and integrate over [0,2]. That's the exact area.
 


The rectangle making part of the Riemann sum has base \Delta x and height equal to the distance between the two curves for some value of x in the interval. Since y= 12- 6x is always above y= x2- 4, that distance is (12- 6x)- (x2- 4)= 16- 6x- x2. The area is the product of those two.

You say the integral is from 0 to 2. I will repeat Dick's question: are you told that, separately in the question? In your first post, you only said that you were finding the area between the graphs: and that runs from x= -8 to x= 2.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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