Mixing
Improper integral - Calculating Area of Floor Function
$begingroup$
I'm stuck trying to calculate the area of the region defined by:
$begin{Bmatrix}
{(x,y)inmathbb{R^2}:0leq xwedge 0leq yleq 2^{-lfloor x rfloor}}
end{Bmatrix}$
I'm just starting studying Calculus I and I don't have much information to give: I could do this exercise if it wasn't for $xleq0$, however I can't calculate the improper integral because I can't integrate the floor function and I though about using an infinite sum (I don't know if that doesn't make sense) but we haven't started learning it yet so I assume that ain't it.
calculus area
asked Jan 23 at 15:20
Mário BelgaMário Belga
103
$endgroup$
add a comment |
$begingroup$
I'm stuck trying to calculate the area of the region defined by:
$begin{Bmatrix}
{(x,y)inmathbb{R^2}:0leq xwedge 0leq yleq 2^{-lfloor x rfloor}}
end{Bmatrix}$
I'm just starting studying Calculus I and I don't have much information to give: I could do this exercise if it wasn't for $xleq0$, however I can't calculate the improper integral because I can't integrate the floor function and I though about using an infinite sum (I don't know if that doesn't make sense) but we haven't started learning it yet so I assume that ain't it.
calculus area
asked Jan 23 at 15:20
Mário BelgaMário Belga
103
$endgroup$
$begingroup$
I guess you mean $(x,y)inmathbb R^2$
$endgroup$
– MPW
Jan 23 at 15:28
$begingroup$
@ajotatxe You're right, I tried to give more information on my situation.
$endgroup$
– Mário Belga
Jan 23 at 16:15
$begingroup$
@MPW Thanks, just fixed it.
$endgroup$
– Mário Belga
Jan 23 at 16:16
add a comment |
$begingroup$
I'm stuck trying to calculate the area of the region defined by:
$begin{Bmatrix}
{(x,y)inmathbb{R^2}:0leq xwedge 0leq yleq 2^{-lfloor x rfloor}}
end{Bmatrix}$
I'm just starting studying Calculus I and I don't have much information to give: I could do this exercise if it wasn't for $xleq0$, however I can't calculate the improper integral because I can't integrate the floor function and I though about using an infinite sum (I don't know if that doesn't make sense) but we haven't started learning it yet so I assume that ain't it.
calculus area
asked Jan 23 at 15:20
Mário BelgaMário Belga
103
$endgroup$
I'm stuck trying to calculate the area of the region defined by:
$begin{Bmatrix}
{(x,y)inmathbb{R^2}:0leq xwedge 0leq yleq 2^{-lfloor x rfloor}}
end{Bmatrix}$
I'm just starting studying Calculus I and I don't have much information to give: I could do this exercise if it wasn't for $xleq0$, however I can't calculate the improper integral because I can't integrate the floor function and I though about using an infinite sum (I don't know if that doesn't make sense) but we haven't started learning it yet so I assume that ain't it.
calculus area
calculus area
asked Jan 23 at 15:20
Mário BelgaMário Belga
103
asked Jan 23 at 15:20
Mário BelgaMário Belga
103
edited Jan 23 at 16:09
Mário Belga
asked Jan 23 at 15:20
Mário BelgaMário Belga
103
asked Jan 23 at 15:20
Mário BelgaMário Belga
103
asked Jan 23 at 15:20
Mário BelgaMário Belga
103
103
$begingroup$
I guess you mean $(x,y)inmathbb R^2$
$endgroup$
– MPW
Jan 23 at 15:28
$begingroup$
@ajotatxe You're right, I tried to give more information on my situation.
$endgroup$
– Mário Belga
Jan 23 at 16:15
$begingroup$
@MPW Thanks, just fixed it.
$endgroup$
– Mário Belga
Jan 23 at 16:16
add a comment |
$begingroup$
I guess you mean $(x,y)inmathbb R^2$
$endgroup$
– MPW
Jan 23 at 15:28
$begingroup$
@ajotatxe You're right, I tried to give more information on my situation.
$endgroup$
– Mário Belga
Jan 23 at 16:15
$begingroup$
@MPW Thanks, just fixed it.
$endgroup$
– Mário Belga
Jan 23 at 16:16
$begingroup$
I guess you mean $(x,y)inmathbb R^2$
$endgroup$
– MPW
Jan 23 at 15:28
$begingroup$
I guess you mean $(x,y)inmathbb R^2$
$endgroup$
– MPW
Jan 23 at 15:28
$begingroup$
@ajotatxe You're right, I tried to give more information on my situation.
$endgroup$
– Mário Belga
Jan 23 at 16:15
$begingroup$
@ajotatxe You're right, I tried to give more information on my situation.
$endgroup$
– Mário Belga
Jan 23 at 16:15
$begingroup$
@MPW Thanks, just fixed it.
$endgroup$
– Mário Belga
Jan 23 at 16:16
$begingroup$
@MPW Thanks, just fixed it.
$endgroup$
– Mário Belga
Jan 23 at 16:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If we plot y against x, $[x]$ is the floor function with steps going upward. However the function $y = 2^{-[x]}$ has steps going downward with each step half of the height before it. Hence if $N <= x < N + 1$ we have area equal to
$$int2^{-[x]}dx = 1 + frac{1}{2} + frac{1}{2^2} + ... + frac{1}{2^{(N - 1)}} + frac{1}{2^N}(x - N)$$
$$= frac{1 - frac{1}{2^N}}{1 - frac{1}{2}} + frac{1}{2^N}(x - N) = 2 - frac{1}{2^N}(2 - (x - N))$$
$$= 2 - frac{1}{2^{[x]}}(2 - x + [x])$$
Since N = [x].
answered Jan 27 at 2:14
KY TangKY Tang
39435
$endgroup$
$begingroup$
To include more than 1 character in superscript, you have to close it using braces. For example, $2^{n-1}$$2^{n-1}$.
$endgroup$
– manooooh
Jan 27 at 3:03
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If we plot y against x, $[x]$ is the floor function with steps going upward. However the function $y = 2^{-[x]}$ has steps going downward with each step half of the height before it. Hence if $N <= x < N + 1$ we have area equal to
$$int2^{-[x]}dx = 1 + frac{1}{2} + frac{1}{2^2} + ... + frac{1}{2^{(N - 1)}} + frac{1}{2^N}(x - N)$$
$$= frac{1 - frac{1}{2^N}}{1 - frac{1}{2}} + frac{1}{2^N}(x - N) = 2 - frac{1}{2^N}(2 - (x - N))$$
$$= 2 - frac{1}{2^{[x]}}(2 - x + [x])$$
Since N = [x].
answered Jan 27 at 2:14
KY TangKY Tang
39435
$endgroup$
$begingroup$
To include more than 1 character in superscript, you have to close it using braces. For example, $2^{n-1}$$2^{n-1}$.
$endgroup$
– manooooh
Jan 27 at 3:03
add a comment |
$begingroup$
If we plot y against x, $[x]$ is the floor function with steps going upward. However the function $y = 2^{-[x]}$ has steps going downward with each step half of the height before it. Hence if $N <= x < N + 1$ we have area equal to
$$int2^{-[x]}dx = 1 + frac{1}{2} + frac{1}{2^2} + ... + frac{1}{2^{(N - 1)}} + frac{1}{2^N}(x - N)$$
$$= frac{1 - frac{1}{2^N}}{1 - frac{1}{2}} + frac{1}{2^N}(x - N) = 2 - frac{1}{2^N}(2 - (x - N))$$
$$= 2 - frac{1}{2^{[x]}}(2 - x + [x])$$
Since N = [x].
answered Jan 27 at 2:14
KY TangKY Tang
39435
$endgroup$
$begingroup$
To include more than 1 character in superscript, you have to close it using braces. For example, $2^{n-1}$$2^{n-1}$.
$endgroup$
– manooooh
Jan 27 at 3:03
add a comment |
$begingroup$
If we plot y against x, $[x]$ is the floor function with steps going upward. However the function $y = 2^{-[x]}$ has steps going downward with each step half of the height before it. Hence if $N <= x < N + 1$ we have area equal to
$$int2^{-[x]}dx = 1 + frac{1}{2} + frac{1}{2^2} + ... + frac{1}{2^{(N - 1)}} + frac{1}{2^N}(x - N)$$
$$= frac{1 - frac{1}{2^N}}{1 - frac{1}{2}} + frac{1}{2^N}(x - N) = 2 - frac{1}{2^N}(2 - (x - N))$$
$$= 2 - frac{1}{2^{[x]}}(2 - x + [x])$$
Since N = [x].
answered Jan 27 at 2:14
KY TangKY Tang
39435
$endgroup$
If we plot y against x, $[x]$ is the floor function with steps going upward. However the function $y = 2^{-[x]}$ has steps going downward with each step half of the height before it. Hence if $N <= x < N + 1$ we have area equal to
$$int2^{-[x]}dx = 1 + frac{1}{2} + frac{1}{2^2} + ... + frac{1}{2^{(N - 1)}} + frac{1}{2^N}(x - N)$$
$$= frac{1 - frac{1}{2^N}}{1 - frac{1}{2}} + frac{1}{2^N}(x - N) = 2 - frac{1}{2^N}(2 - (x - N))$$
$$= 2 - frac{1}{2^{[x]}}(2 - x + [x])$$
Since N = [x].
answered Jan 27 at 2:14
KY TangKY Tang
39435
edited Jan 27 at 3:18
answered Jan 27 at 2:14
KY TangKY Tang
39435
answered Jan 27 at 2:14
KY TangKY Tang
39435
answered Jan 27 at 2:14
KY TangKY Tang
39435
39435
$begingroup$
To include more than 1 character in superscript, you have to close it using braces. For example, $2^{n-1}$$2^{n-1}$.
$endgroup$
– manooooh
Jan 27 at 3:03
add a comment |
$begingroup$
To include more than 1 character in superscript, you have to close it using braces. For example, $2^{n-1}$$2^{n-1}$.
$endgroup$
– manooooh
Jan 27 at 3:03
$begingroup$
To include more than 1 character in superscript, you have to close it using braces. For example, $2^{n-1}$
$2^{n-1}$.$endgroup$
– manooooh
Jan 27 at 3:03
$begingroup$
To include more than 1 character in superscript, you have to close it using braces. For example, $2^{n-1}$
$2^{n-1}$.$endgroup$
– manooooh
Jan 27 at 3:03
add a comment |
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$begingroup$
I guess you mean $(x,y)inmathbb R^2$
$endgroup$
– MPW
Jan 23 at 15:28
$begingroup$
@ajotatxe You're right, I tried to give more information on my situation.
$endgroup$
– Mário Belga
Jan 23 at 16:15
$begingroup$
@MPW Thanks, just fixed it.
$endgroup$
– Mário Belga
Jan 23 at 16:16