Why is the area from $1$ to $2$ of $frac{1}{x}$ the same as the area from $6$ to $12$?
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I get this in the calculus sense- you integrate and $ln frac{2}{1}$ and $ln frac{12}{6}$ are the same, but how would I answer this with a non-calculus answer? Why do they have the same area?
calculus
edited Jan 16 at 23:57
whiskeyo
1368
asked Jan 16 at 23:04
user312492user312492
6
$endgroup$
add a comment |
$begingroup$
I get this in the calculus sense- you integrate and $ln frac{2}{1}$ and $ln frac{12}{6}$ are the same, but how would I answer this with a non-calculus answer? Why do they have the same area?
calculus
edited Jan 16 at 23:57
whiskeyo
1368
asked Jan 16 at 23:04
user312492user312492
6
$endgroup$
4
$begingroup$
Not in a rigorous way: the transformation $x=6X$ and $y=Y/6$ preserves the area of rectangles, so it preserves areas. The curve $xy=1$ is mapped to $XY=1$ and the trapezoid defining the area from $6$ to $12$ is mapped to the trapezoid defining the area from $1$ to $2$.
$endgroup$
– egreg
Jan 17 at 0:00
$begingroup$
^oughta be an answer.
$endgroup$
– Randall
Jan 17 at 2:28
add a comment |
$begingroup$
I get this in the calculus sense- you integrate and $ln frac{2}{1}$ and $ln frac{12}{6}$ are the same, but how would I answer this with a non-calculus answer? Why do they have the same area?
calculus
edited Jan 16 at 23:57
whiskeyo
1368
asked Jan 16 at 23:04
user312492user312492
6
$endgroup$
I get this in the calculus sense- you integrate and $ln frac{2}{1}$ and $ln frac{12}{6}$ are the same, but how would I answer this with a non-calculus answer? Why do they have the same area?
calculus
calculus
edited Jan 16 at 23:57
whiskeyo
1368
asked Jan 16 at 23:04
user312492user312492
6
edited Jan 16 at 23:57
whiskeyo
1368
asked Jan 16 at 23:04
user312492user312492
6
edited Jan 16 at 23:57
whiskeyo
1368
edited Jan 16 at 23:57
whiskeyo
1368
edited Jan 16 at 23:57
whiskeyo
1368
1368
asked Jan 16 at 23:04
user312492user312492
6
asked Jan 16 at 23:04
user312492user312492
6
asked Jan 16 at 23:04
user312492user312492
6
6
4
$begingroup$
Not in a rigorous way: the transformation $x=6X$ and $y=Y/6$ preserves the area of rectangles, so it preserves areas. The curve $xy=1$ is mapped to $XY=1$ and the trapezoid defining the area from $6$ to $12$ is mapped to the trapezoid defining the area from $1$ to $2$.
$endgroup$
– egreg
Jan 17 at 0:00
$begingroup$
^oughta be an answer.
$endgroup$
– Randall
Jan 17 at 2:28
add a comment |
4
$begingroup$
Not in a rigorous way: the transformation $x=6X$ and $y=Y/6$ preserves the area of rectangles, so it preserves areas. The curve $xy=1$ is mapped to $XY=1$ and the trapezoid defining the area from $6$ to $12$ is mapped to the trapezoid defining the area from $1$ to $2$.
$endgroup$
– egreg
Jan 17 at 0:00
$begingroup$
^oughta be an answer.
$endgroup$
– Randall
Jan 17 at 2:28
4
4
$begingroup$
Not in a rigorous way: the transformation $x=6X$ and $y=Y/6$ preserves the area of rectangles, so it preserves areas. The curve $xy=1$ is mapped to $XY=1$ and the trapezoid defining the area from $6$ to $12$ is mapped to the trapezoid defining the area from $1$ to $2$.
$endgroup$
– egreg
Jan 17 at 0:00
$begingroup$
Not in a rigorous way: the transformation $x=6X$ and $y=Y/6$ preserves the area of rectangles, so it preserves areas. The curve $xy=1$ is mapped to $XY=1$ and the trapezoid defining the area from $6$ to $12$ is mapped to the trapezoid defining the area from $1$ to $2$.
$endgroup$
– egreg
Jan 17 at 0:00
$begingroup$
^oughta be an answer.
$endgroup$
– Randall
Jan 17 at 2:28
$begingroup$
^oughta be an answer.
$endgroup$
– Randall
Jan 17 at 2:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let
$$
I = int_{x=1}^{x=2} frac{dx}{x}.
$$
Multiplying and dividing this expression by $6$, we have
$$
I = int_{x=1}^{x=2} frac{d(6x)}{6x}.
$$
Using the substitution $y=6x$, we have $dy=6dx$. When $x=1$, $y=6$, and when $x=2$, $y=12$. Therefore,
$$I = int_{y=6}^{y=12} frac{dy}{y}.$$
answered Jan 16 at 23:11
rafa11111rafa11111
1,2042417
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add a comment |
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1 Answer
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$begingroup$
Let
$$
I = int_{x=1}^{x=2} frac{dx}{x}.
$$
Multiplying and dividing this expression by $6$, we have
$$
I = int_{x=1}^{x=2} frac{d(6x)}{6x}.
$$
Using the substitution $y=6x$, we have $dy=6dx$. When $x=1$, $y=6$, and when $x=2$, $y=12$. Therefore,
$$I = int_{y=6}^{y=12} frac{dy}{y}.$$
answered Jan 16 at 23:11
rafa11111rafa11111
1,2042417
$endgroup$
add a comment |
$begingroup$
Let
$$
I = int_{x=1}^{x=2} frac{dx}{x}.
$$
Multiplying and dividing this expression by $6$, we have
$$
I = int_{x=1}^{x=2} frac{d(6x)}{6x}.
$$
Using the substitution $y=6x$, we have $dy=6dx$. When $x=1$, $y=6$, and when $x=2$, $y=12$. Therefore,
$$I = int_{y=6}^{y=12} frac{dy}{y}.$$
answered Jan 16 at 23:11
rafa11111rafa11111
1,2042417
$endgroup$
add a comment |
$begingroup$
Let
$$
I = int_{x=1}^{x=2} frac{dx}{x}.
$$
Multiplying and dividing this expression by $6$, we have
$$
I = int_{x=1}^{x=2} frac{d(6x)}{6x}.
$$
Using the substitution $y=6x$, we have $dy=6dx$. When $x=1$, $y=6$, and when $x=2$, $y=12$. Therefore,
$$I = int_{y=6}^{y=12} frac{dy}{y}.$$
answered Jan 16 at 23:11
rafa11111rafa11111
1,2042417
$endgroup$
Let
$$
I = int_{x=1}^{x=2} frac{dx}{x}.
$$
Multiplying and dividing this expression by $6$, we have
$$
I = int_{x=1}^{x=2} frac{d(6x)}{6x}.
$$
Using the substitution $y=6x$, we have $dy=6dx$. When $x=1$, $y=6$, and when $x=2$, $y=12$. Therefore,
$$I = int_{y=6}^{y=12} frac{dy}{y}.$$
answered Jan 16 at 23:11
rafa11111rafa11111
1,2042417
answered Jan 16 at 23:11
rafa11111rafa11111
1,2042417
answered Jan 16 at 23:11
rafa11111rafa11111
1,2042417
answered Jan 16 at 23:11
rafa11111rafa11111
1,2042417
1,2042417
add a comment |
add a comment |
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4
$begingroup$
Not in a rigorous way: the transformation $x=6X$ and $y=Y/6$ preserves the area of rectangles, so it preserves areas. The curve $xy=1$ is mapped to $XY=1$ and the trapezoid defining the area from $6$ to $12$ is mapped to the trapezoid defining the area from $1$ to $2$.
$endgroup$
– egreg
Jan 17 at 0:00
$begingroup$
^oughta be an answer.
$endgroup$
– Randall
Jan 17 at 2:28