Mixing
Why did we use 2 region of integration? [closed]
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In this case, why did we use two region of integration although x/2 is greater than y only in one case.
Note that initially, y is between 0 and 1 and that x is between square root of y and 2.
integration
asked Jul 22 at 20:58
Roy Rizk
887
closed as unclear what you're asking by m_t_, Did, Xander Henderson, Leucippus, Taroccoesbrocco Jul 23 at 6:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-3
down vote
favorite
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In this case, why did we use two region of integration although x/2 is greater than y only in one case.
Note that initially, y is between 0 and 1 and that x is between square root of y and 2.
integration
asked Jul 22 at 20:58
Roy Rizk
887
closed as unclear what you're asking by m_t_, Did, Xander Henderson, Leucippus, Taroccoesbrocco Jul 23 at 6:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
Have you drawn the region?
– G Tony Jacobs
Jul 22 at 21:04
add a comment |Â
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
.
In this case, why did we use two region of integration although x/2 is greater than y only in one case.
Note that initially, y is between 0 and 1 and that x is between square root of y and 2.
integration
asked Jul 22 at 20:58
Roy Rizk
887
.
In this case, why did we use two region of integration although x/2 is greater than y only in one case.
Note that initially, y is between 0 and 1 and that x is between square root of y and 2.
integration
asked Jul 22 at 20:58
Roy Rizk
887
asked Jul 22 at 20:58
Roy Rizk
887
asked Jul 22 at 20:58
Roy Rizk
887
asked Jul 22 at 20:58
Roy Rizk
887
887
closed as unclear what you're asking by m_t_, Did, Xander Henderson, Leucippus, Taroccoesbrocco Jul 23 at 6:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by m_t_, Did, Xander Henderson, Leucippus, Taroccoesbrocco Jul 23 at 6:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
Have you drawn the region?
– G Tony Jacobs
Jul 22 at 21:04
add a comment |Â
2
Have you drawn the region?
– G Tony Jacobs
Jul 22 at 21:04
2
2
Have you drawn the region?
– G Tony Jacobs
Jul 22 at 21:04
Have you drawn the region?
– G Tony Jacobs
Jul 22 at 21:04
add a comment |Â
1 Answer
1
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votes
up vote
1
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accepted
Perhaps this picture makes it clear. The region beneath $y=frac14$ has a different left boundary from the region above that line. Algebraically, $sqrty>2y$ when $y<frac14$, but not when $y>frac14$
This is the intersection of the two regions $sqrty<x<2$ (from the original setup) and $y<fracx2$ (from the probability we're looking for now).
answered Jul 22 at 21:33
G Tony Jacobs
25.6k43483
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Perhaps this picture makes it clear. The region beneath $y=frac14$ has a different left boundary from the region above that line. Algebraically, $sqrty>2y$ when $y<frac14$, but not when $y>frac14$
This is the intersection of the two regions $sqrty<x<2$ (from the original setup) and $y<fracx2$ (from the probability we're looking for now).
answered Jul 22 at 21:33
G Tony Jacobs
25.6k43483
add a comment |Â
up vote
1
down vote
accepted
Perhaps this picture makes it clear. The region beneath $y=frac14$ has a different left boundary from the region above that line. Algebraically, $sqrty>2y$ when $y<frac14$, but not when $y>frac14$
This is the intersection of the two regions $sqrty<x<2$ (from the original setup) and $y<fracx2$ (from the probability we're looking for now).
answered Jul 22 at 21:33
G Tony Jacobs
25.6k43483
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Perhaps this picture makes it clear. The region beneath $y=frac14$ has a different left boundary from the region above that line. Algebraically, $sqrty>2y$ when $y<frac14$, but not when $y>frac14$
This is the intersection of the two regions $sqrty<x<2$ (from the original setup) and $y<fracx2$ (from the probability we're looking for now).
answered Jul 22 at 21:33
G Tony Jacobs
25.6k43483
Perhaps this picture makes it clear. The region beneath $y=frac14$ has a different left boundary from the region above that line. Algebraically, $sqrty>2y$ when $y<frac14$, but not when $y>frac14$
This is the intersection of the two regions $sqrty<x<2$ (from the original setup) and $y<fracx2$ (from the probability we're looking for now).
answered Jul 22 at 21:33
G Tony Jacobs
25.6k43483
answered Jul 22 at 21:33
G Tony Jacobs
25.6k43483
answered Jul 22 at 21:33
G Tony Jacobs
25.6k43483
answered Jul 22 at 21:33
G Tony Jacobs
25.6k43483
25.6k43483
add a comment |Â
add a comment |Â
2
Have you drawn the region?
– G Tony Jacobs
Jul 22 at 21:04