Finding the domain of an Integral
I'm trying to figure out what the domain of this function is.
I was thinking that the domain could be the maximum and minimum amounts of area that is accumulated, but I feel like that's the range of the function rather than the domain. Could anyone point me in the right direction to solve this question?
integration definite-integrals
edited Dec 3 '18 at 9:31
Emilio Novati
51.5k43472
asked Dec 3 '18 at 9:00
kydd
197
add a comment |
I'm trying to figure out what the domain of this function is.
I was thinking that the domain could be the maximum and minimum amounts of area that is accumulated, but I feel like that's the range of the function rather than the domain. Could anyone point me in the right direction to solve this question?
integration definite-integrals
edited Dec 3 '18 at 9:31
Emilio Novati
51.5k43472
asked Dec 3 '18 at 9:00
kydd
197
add a comment |
I'm trying to figure out what the domain of this function is.
I was thinking that the domain could be the maximum and minimum amounts of area that is accumulated, but I feel like that's the range of the function rather than the domain. Could anyone point me in the right direction to solve this question?
integration definite-integrals
edited Dec 3 '18 at 9:31
Emilio Novati
51.5k43472
asked Dec 3 '18 at 9:00
kydd
197
I'm trying to figure out what the domain of this function is.
I was thinking that the domain could be the maximum and minimum amounts of area that is accumulated, but I feel like that's the range of the function rather than the domain. Could anyone point me in the right direction to solve this question?
integration definite-integrals
integration definite-integrals
edited Dec 3 '18 at 9:31
Emilio Novati
51.5k43472
asked Dec 3 '18 at 9:00
kydd
197
edited Dec 3 '18 at 9:31
Emilio Novati
51.5k43472
asked Dec 3 '18 at 9:00
kydd
197
edited Dec 3 '18 at 9:31
Emilio Novati
51.5k43472
edited Dec 3 '18 at 9:31
Emilio Novati
51.5k43472
edited Dec 3 '18 at 9:31
Emilio Novati
51.5k43472
51.5k43472
asked Dec 3 '18 at 9:00
kydd
197
asked Dec 3 '18 at 9:00
kydd
197
asked Dec 3 '18 at 9:00
kydd
197
197
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
From the figure we see that the limit of integration is:
$$
0le 2x-1 le 6
$$
because the function $f(t)$ is:
$$
f(t)=begin{cases}
sqrt{4-(t-2)^2} quad for quad 0le t le 4\
sqrt{}1-(t-5)^2 quad for quad 4le t le 6
end{cases}
$$
from which you can find the domain of the function: $[frac{1}{2},frac{7}{2}]$.
And , as you say, the range is the interval between the minimum and maximum values of the area: $[0,2pi]$.
answered Dec 3 '18 at 9:39
Emilio Novati
51.5k43472
The minimum value of $h(x)=0$ at $x=1/2$
– Shubham Johri
Dec 3 '18 at 10:00
Yes, ...My stupid mistake! Thank you :)
– Emilio Novati
Dec 3 '18 at 10:04
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
From the figure we see that the limit of integration is:
$$
0le 2x-1 le 6
$$
because the function $f(t)$ is:
$$
f(t)=begin{cases}
sqrt{4-(t-2)^2} quad for quad 0le t le 4\
sqrt{}1-(t-5)^2 quad for quad 4le t le 6
end{cases}
$$
from which you can find the domain of the function: $[frac{1}{2},frac{7}{2}]$.
And , as you say, the range is the interval between the minimum and maximum values of the area: $[0,2pi]$.
answered Dec 3 '18 at 9:39
Emilio Novati
51.5k43472
The minimum value of $h(x)=0$ at $x=1/2$
– Shubham Johri
Dec 3 '18 at 10:00
Yes, ...My stupid mistake! Thank you :)
– Emilio Novati
Dec 3 '18 at 10:04
add a comment |
From the figure we see that the limit of integration is:
$$
0le 2x-1 le 6
$$
because the function $f(t)$ is:
$$
f(t)=begin{cases}
sqrt{4-(t-2)^2} quad for quad 0le t le 4\
sqrt{}1-(t-5)^2 quad for quad 4le t le 6
end{cases}
$$
from which you can find the domain of the function: $[frac{1}{2},frac{7}{2}]$.
And , as you say, the range is the interval between the minimum and maximum values of the area: $[0,2pi]$.
answered Dec 3 '18 at 9:39
Emilio Novati
51.5k43472
The minimum value of $h(x)=0$ at $x=1/2$
– Shubham Johri
Dec 3 '18 at 10:00
Yes, ...My stupid mistake! Thank you :)
– Emilio Novati
Dec 3 '18 at 10:04
add a comment |
From the figure we see that the limit of integration is:
$$
0le 2x-1 le 6
$$
because the function $f(t)$ is:
$$
f(t)=begin{cases}
sqrt{4-(t-2)^2} quad for quad 0le t le 4\
sqrt{}1-(t-5)^2 quad for quad 4le t le 6
end{cases}
$$
from which you can find the domain of the function: $[frac{1}{2},frac{7}{2}]$.
And , as you say, the range is the interval between the minimum and maximum values of the area: $[0,2pi]$.
answered Dec 3 '18 at 9:39
Emilio Novati
51.5k43472
From the figure we see that the limit of integration is:
$$
0le 2x-1 le 6
$$
because the function $f(t)$ is:
$$
f(t)=begin{cases}
sqrt{4-(t-2)^2} quad for quad 0le t le 4\
sqrt{}1-(t-5)^2 quad for quad 4le t le 6
end{cases}
$$
from which you can find the domain of the function: $[frac{1}{2},frac{7}{2}]$.
And , as you say, the range is the interval between the minimum and maximum values of the area: $[0,2pi]$.
answered Dec 3 '18 at 9:39
Emilio Novati
51.5k43472
edited Dec 3 '18 at 10:03
answered Dec 3 '18 at 9:39
Emilio Novati
51.5k43472
answered Dec 3 '18 at 9:39
Emilio Novati
51.5k43472
answered Dec 3 '18 at 9:39
Emilio Novati
51.5k43472
51.5k43472
The minimum value of $h(x)=0$ at $x=1/2$
– Shubham Johri
Dec 3 '18 at 10:00
Yes, ...My stupid mistake! Thank you :)
– Emilio Novati
Dec 3 '18 at 10:04
add a comment |
The minimum value of $h(x)=0$ at $x=1/2$
– Shubham Johri
Dec 3 '18 at 10:00
Yes, ...My stupid mistake! Thank you :)
– Emilio Novati
Dec 3 '18 at 10:04
The minimum value of $h(x)=0$ at $x=1/2$
– Shubham Johri
Dec 3 '18 at 10:00
The minimum value of $h(x)=0$ at $x=1/2$
– Shubham Johri
Dec 3 '18 at 10:00
Yes, ...My stupid mistake! Thank you :)
– Emilio Novati
Dec 3 '18 at 10:04
Yes, ...My stupid mistake! Thank you :)
– Emilio Novati
Dec 3 '18 at 10:04
add a comment |
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