course Mth 272 ’BË€ÿìöÈÀJŒ½ÐÏoâ·õ„ô™ÏÌïÉÚô¶Âºassignment #032
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09:01:05 Query problem 7.8.6 integrate (x^2+y^2) with respect to y from x^2 to `sqrt(x)
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RESPONSE --> you want to hold x constant first you need to integrate (x^2+y^2) = y^3 / 3 [y^3 / 3] from x^2 to 'sqrt(x) then apply fundamental theorem ['sqrt(x)^3 / 3] - [(x^2)^3 / 3] confidence assessment: 2
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09:01:30 What is your result?
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RESPONSE --> previously shown confidence assessment: 2
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09:01:55 What was your antiderivative?
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RESPONSE --> y^3 / 3 confidence assessment: 2
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09:07:51 Query problem 7.8.14 integrate `sqrt(1-x^2) from 0 to x wrt y then from 0 to 1 wrt x
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RESPONSE --> wrt y `sqrt(1-x^2) becomes [y] apply fundamental theorem [x] - [0] = x find antiderivative of x = [x^2 / 2] from 0 to 1 = [1^2 / 2] - [0^2 / 2] = (1/2) - (0) = 1/2 confidence assessment: 2
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09:08:17 With respect to which variable of integration did you first integrate?
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RESPONSE --> with respect to y confidence assessment: 2
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09:10:41 What was the antiderivative of your first integration and what was the definite integral you obtained when you substituted the limits?
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RESPONSE --> [y] was the antiderivative substituting the limits was [x] - [0] = x confidence assessment: 2
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09:10:51 What did you get when you integrated this expression with respect to the remaining variable of integration?
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RESPONSE --> previously answered confidence assessment:
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09:28:01 Query problem 7.8.32 sketch region and reverse order of integration for integral of 1 from 0 to 4-y^2, then from -2 to 2.
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RESPONSE --> from the limits of integration i know that 0 < x < 4 - y^2 which means that the region R is bounded on the left by the parabola the line x = 0 and the parabola 4 -y^2 and because -2 < x < 2, I know that the region lies both above and below the x-axis. If you reverse the order of integration -2 < x < 0 then solving for x = 4 - y^2 bounds for y are 0 < y < 'sqrt(4x) [4 - y^2] - [0] = 4 - y^2 find antiderivative = -2y from -2 to 2 [-2(-2)] - [-2(2)] = 4 - (-4) = 8 and reversing order [0] - [-2] = 2 antiderivative of 2 = 2x from 0 to 'sqrt(4x) [2('sqrt(4x)] - [0] = 4x hmm, i know these are supposed to be the same, but obviously I went wrong somewhere. confidence assessment: 1
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09:28:10 What are the limits, inner limit first, when the order of integration was changed?
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RESPONSE --> previously shown confidence assessment:
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09:28:21 Describe the region in the xy plane whose area is given by the given integral.
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RESPONSE --> previously answered confidence assessment:
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09:28:30 Explain how you obtained the limits for the integral when the order of integration was changed.
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RESPONSE --> previously answered confidence assessment:
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09:28:38 What did you get when you evaluated the original integral?
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RESPONSE --> previously answered confidence assessment:
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09:28:47 What did you get when you evaluated the integral with the order of integration reversed?
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RESPONSE --> previously answered confidence assessment:
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09:29:32 Why should both integrals give you the same result?
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RESPONSE --> because they are both quantifying the same area of Region R confidence assessment: 2
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09:38:55 Query problem 7.8.40 Area beneath curve 1/`sqrt(x-1) for 2 <= x <= 5
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RESPONSE --> you're going to use a double integral here. Im not sure what one of the bounds will be. i know one is from 2 to 5. it doesn't give you the other bound, It might be from 1 to y integrate with respect to x = (x-1)^1/2 / 1/2 apply fundamental theorem from 1 to y [(y-1)^1/2 / 1/2] - [(1-1)^1/2 / 1/2] I can tell that this is not right with bound from 1 to y, but I don't see where you get the other bound from. If I knew the other bound then I would integrate with respecto to x here, then apply the fundamental theorem. Then I would integrate that answer and apply the fundamental theorem from 2 to 5. confidence assessment: 2
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09:39:07 What are the limits and the order of integration for the double integral you used to find the area?
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RESPONSE --> previously answered confidence assessment:
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09:39:40 What is the area?
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RESPONSE --> previously answered the answer would be in square units though. confidence assessment:
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09:43:11 Query problem 7.8.44 area bounded by xy=9, y=x, y=0, x=9
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RESPONSE --> I know you want to begin by sketching the region R to find the bounds. I don't know how to sketch it to get those bounds. confidence assessment:
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09:43:17 What is the area of the region?
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RESPONSE --> confidence assessment:
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09:43:25 Describe the region in detail.
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RESPONSE --> confidence assessment:
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09:43:54 Did you have to use one double integral or two to find the area?
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RESPONSE --> probably had to use one double integral confidence assessment: 1
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09:44:01 What was the order of integration and what were the limits you used for each integral?
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RESPONSE --> confidence assessment:
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