Query Questions 122

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course Mth 277

November 28 around 11:20pm.

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Question: `q001. Evaluate the double integral Int( Int((x^2* e^xy) dy), 0, x) dx, 0 , 1)

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Your solution:

Int (x^2*e^(xy) from 0 to x) dy

xe^(xy) from 0 to x gives me x(e^(x^2)-1)

Int (x(e^(x^2)-1) from 0 to 1) dx

(e^(x^2)/2)-(x^2/2) from 0 to 1 gives me (e/2)-1

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Given Solution:

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Self-critique (if necessary):

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Question: `q002. Integrate 4x over D with respect to A using a double integral where D is the region bounded by 4 - x^2, y = 3x, and x = 0.

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Your solution:

0<=x<=1 and 3x<=y<=4-x^2

Int (4x dy) from 3x to 4-x^2

4x from 3x to 4-x^2 gives me [4(4-x^2)]-[4(3x)] OR -4x^2-12x+16

Int (-4x^2-12x+16 dx) from 0 to 1

(-4x^3/3)-6x^2+16x from 0 to 1 gives me 26/3

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Given Solution:

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Self-critique (if necessary):

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Question: `q003. Compute the following integral with the given order of integration and with the order changed: Int[ Int(4x^3 dy, 0, sqrt(x)) dx, 0, 4].

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Your solution:

Int (4x^3 dy) from 0 to sqrt x

=4x^(7/2)

Int (4x^(7/2) dx) from 0 to 4

=(8x^(9/2))/9 from 0 to 4

=4096/9 or 455.11

It’s asking to change the order and do it again and I don’t know how to do that. I DO think I would have to graph it and looking at the graph, see the regions that will correspond to it.

Check the Query document solutions. y = sqrt(x) is the bounding curve; should be easy to graph. That corresponds to x = y^2, also easy to graph.

For a given y between 0 and 2, the graph goes from x = y^2 to x = 4.

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Given Solution:

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Self-critique (if necessary):

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self-critique rating #$&*:

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Question: `q004. Give two different ways to set up the integral of the area of the region D where D is the region in the first quadrant of the xy-plane bounded by y = 4/x^2 and y = 5 - x^2. Evaluate one of these integrals.

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Your solution:

I got caught up on this one but I’ll put down what I attempted and maybe you can help me go from there:

(4/x^2)=(5-x^2)

4=5x^2-x^4

-x^4+5x^2-4=0

x= +-1, +-2

A= (Int Int (dy, (4/x^2), (5-x^2)) dx, 1, 2)

=Int ((5-x^2)-(4/x^2) dx)

=(-x^3/3)+5x+(4/x) from 1 to 2

=2/3

This looks good. Check the Query document for reversal of the integral.

Solutions in the first quadrant for x are x = sqrt(4/y) and x = sqrt(y - 5). y goes from 0 to 4, x from sqrt(4/y) to sqrt(y - 5).

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Given Solution:

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Self-critique (if necessary):"

Check against my notes, as especially the solutions in the Query, some of which include graphs.