Hello! I have this exercise with its answer. Before the answer was displayed, I could not see the graphic. My question is, how can I determine the lower limit if I am not looking at the graphic.? It showed there, but I did not get it. Thank you. Find the area of the region enclosed by y = Væ + 2, and y = 5. Algebraically find the limits ofintegration and round your limits of integration and answer to 2 decimal places.The area of the encloses a region is 1x o square units.Show Detailed Solution2 4 6 8 10 12 14 16 18 20 22 24We need to find the lower and upper limits: We need to find the lower and upper limits:The lower limit: x + 2 implies the + 2 > 0lower limit = – 2The upper limit: 5Vz + 2(5) = x + 225 – 2 = xUpper limit = 23%3D23(5 – Væ + 2)-23(x + 2):5x3232(x + 2)5x3- 22(23 + 2)2( – 2 + 2)ž5(23) -5( – 2) -332(25)2(0)- 1011533(31.666667) – ( – 10)41.672.

Since region is enclosed by y=x+2 & y=5. It can be seen clearly that y=5 is upper bound , that…

Answered: Hello! I have this exercise with its…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Hello! I have this exercise with its answer. Before the answer was displayed, I could not see the graphic. My question is, how can I determine the lower limit if I am not looking at the graphic.? It showed there, but I did not get it. Thank you.

Find the area of the region enclosed by y = Væ + 2, and y = 5. Algebraically find the limits of
integration and round your limits of integration and answer to 2 decimal places.
The area of the encloses a region is 1
x o square units.
Show Detailed Solution
2 4 6 8 10 12 14 16 18 20 22 24
We need to find the lower and upper limits:

We need to find the lower and upper limits:
The lower limit: x + 2 implies the + 2 > 0
lower limit = – 2
The upper limit: 5
Vz + 2
(5) = x + 2
25 – 2 = x
Upper limit = 23
%3D
23
(5 – Væ + 2)
-
23
(x + 2):
5x
3
23
2(x + 2)
5x
3
- 2
2(23 + 2)
2( – 2 + 2)ž
5(23) -
5( – 2) -
3
3
2(25)
2(0)
- 10
115
3
3
(31.666667) – ( – 10)
41.67
2.

Expert Solution

Step 1

Since region is enclosed by $y = \sqrt{x + 2} & y = 5$ .

It can be seen clearly that y=5 is upper bound , that is

region is bounded above by y=5.

Now , at y-axis the value of x is zero.

So Region is bounded below by $y = \sqrt{2}$ .

This is the limits when you approach through y-axis.

Now , if you approach through x-axis then then limit will be

$5 = \sqrt{x + 2}$

It implies region is bounded above by x=23.

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