Mixing
Integration of a function and its inverse
$begingroup$
Let $g(x)$ be the inverse function of the function $f(x)$ where
$xin [a,b], f(a) = c, f(b) = d$.
Find the value of
$$int_{a}^{b} f(x) dx - int_{c}^{d} g(y) dy$$
My turn:
$$int_{a}^{b} f(x) dx = F(b) -F(a)$$ where $F'(x) = f(x)$. And
$$int_{c}^{d} g(y) dy = G(d) - G(c) = G(f(b)) - G(f(a))$$
where $G'(y) = g(y)$.
But I did not find a relation between the anti-derivative of f and the anti-derivative of its inverse.
calculus integration
edited Jan 26 at 17:28
Hanno
2,469628
asked Jan 26 at 16:41
Hussien MohamedHussien Mohamed
244
$endgroup$
add a comment |
$begingroup$
Let $g(x)$ be the inverse function of the function $f(x)$ where
$xin [a,b], f(a) = c, f(b) = d$.
Find the value of
$$int_{a}^{b} f(x) dx - int_{c}^{d} g(y) dy$$
My turn:
$$int_{a}^{b} f(x) dx = F(b) -F(a)$$ where $F'(x) = f(x)$. And
$$int_{c}^{d} g(y) dy = G(d) - G(c) = G(f(b)) - G(f(a))$$
where $G'(y) = g(y)$.
But I did not find a relation between the anti-derivative of f and the anti-derivative of its inverse.
calculus integration
edited Jan 26 at 17:28
Hanno
2,469628
asked Jan 26 at 16:41
Hussien MohamedHussien Mohamed
244
$endgroup$
add a comment |
$begingroup$
Let $g(x)$ be the inverse function of the function $f(x)$ where
$xin [a,b], f(a) = c, f(b) = d$.
Find the value of
$$int_{a}^{b} f(x) dx - int_{c}^{d} g(y) dy$$
My turn:
$$int_{a}^{b} f(x) dx = F(b) -F(a)$$ where $F'(x) = f(x)$. And
$$int_{c}^{d} g(y) dy = G(d) - G(c) = G(f(b)) - G(f(a))$$
where $G'(y) = g(y)$.
But I did not find a relation between the anti-derivative of f and the anti-derivative of its inverse.
calculus integration
edited Jan 26 at 17:28
Hanno
2,469628
asked Jan 26 at 16:41
Hussien MohamedHussien Mohamed
244
$endgroup$
Let $g(x)$ be the inverse function of the function $f(x)$ where
$xin [a,b], f(a) = c, f(b) = d$.
Find the value of
$$int_{a}^{b} f(x) dx - int_{c}^{d} g(y) dy$$
My turn:
$$int_{a}^{b} f(x) dx = F(b) -F(a)$$ where $F'(x) = f(x)$. And
$$int_{c}^{d} g(y) dy = G(d) - G(c) = G(f(b)) - G(f(a))$$
where $G'(y) = g(y)$.
But I did not find a relation between the anti-derivative of f and the anti-derivative of its inverse.
calculus integration
calculus integration
edited Jan 26 at 17:28
Hanno
2,469628
asked Jan 26 at 16:41
Hussien MohamedHussien Mohamed
244
edited Jan 26 at 17:28
Hanno
2,469628
asked Jan 26 at 16:41
Hussien MohamedHussien Mohamed
244
edited Jan 26 at 17:28
Hanno
2,469628
edited Jan 26 at 17:28
Hanno
2,469628
edited Jan 26 at 17:28
Hanno
2,469628
2,469628
asked Jan 26 at 16:41
Hussien MohamedHussien Mohamed
244
asked Jan 26 at 16:41
Hussien MohamedHussien Mohamed
244
asked Jan 26 at 16:41
Hussien MohamedHussien Mohamed
244
244
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For brevity, let's consider a special case: $f: [0,b] to [0,d]$ increasing, continuous, onto. If you draw a picture then you realize that the first integral is the area of the rectangle $R:=[0,b] times [0,d]$ under the graph of $f$ and the second integral is the area of $R$ above the graph - that's because the graph of the inverse is symmetrical about the diagonal and the second integral is the area under this graph. But if you reflect it around the diagonal, you get the graph of the original $f$ and you count the area "to the left" from it.
- So if you add them together, not subtract, then you get the area of
$R$, i.e. $bd$. In general situation, you have to take into account
more rectangles but it should be easily deducible from the picture of
the situation. - In your case, when you subtract the second integral from the first one, you can get pretty much everything
between $-bd$ and $bd$. Just consider functions $x^n$ and inverses
$sqrt[n]{x}$ on $[0,1]$.
answered Jan 27 at 12:45
Roman HricRoman Hric
32115
$endgroup$
$begingroup$
Do you it does not have a certain answer ?
$endgroup$
– Hussien Mohamed
Jan 27 at 20:18
$begingroup$
As I wrote in the answer, take e.g. $f(x) = x^n$ on $[0,1]$, then its inverse is $g(y) = sqrt[n]{y}$ on $[0,1]$, and we have $int_{0}^{1} x^n , dx - int_{0}^{1} sqrt[n]{y} , dy = 2/n -1$.
$endgroup$
– Roman Hric
Jan 28 at 17:14
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
$begingroup$
For brevity, let's consider a special case: $f: [0,b] to [0,d]$ increasing, continuous, onto. If you draw a picture then you realize that the first integral is the area of the rectangle $R:=[0,b] times [0,d]$ under the graph of $f$ and the second integral is the area of $R$ above the graph - that's because the graph of the inverse is symmetrical about the diagonal and the second integral is the area under this graph. But if you reflect it around the diagonal, you get the graph of the original $f$ and you count the area "to the left" from it.
- So if you add them together, not subtract, then you get the area of
$R$, i.e. $bd$. In general situation, you have to take into account
more rectangles but it should be easily deducible from the picture of
the situation. - In your case, when you subtract the second integral from the first one, you can get pretty much everything
between $-bd$ and $bd$. Just consider functions $x^n$ and inverses
$sqrt[n]{x}$ on $[0,1]$.
answered Jan 27 at 12:45
Roman HricRoman Hric
32115
$endgroup$
$begingroup$
Do you it does not have a certain answer ?
$endgroup$
– Hussien Mohamed
Jan 27 at 20:18
$begingroup$
As I wrote in the answer, take e.g. $f(x) = x^n$ on $[0,1]$, then its inverse is $g(y) = sqrt[n]{y}$ on $[0,1]$, and we have $int_{0}^{1} x^n , dx - int_{0}^{1} sqrt[n]{y} , dy = 2/n -1$.
$endgroup$
– Roman Hric
Jan 28 at 17:14
add a comment |
$begingroup$
For brevity, let's consider a special case: $f: [0,b] to [0,d]$ increasing, continuous, onto. If you draw a picture then you realize that the first integral is the area of the rectangle $R:=[0,b] times [0,d]$ under the graph of $f$ and the second integral is the area of $R$ above the graph - that's because the graph of the inverse is symmetrical about the diagonal and the second integral is the area under this graph. But if you reflect it around the diagonal, you get the graph of the original $f$ and you count the area "to the left" from it.
- So if you add them together, not subtract, then you get the area of
$R$, i.e. $bd$. In general situation, you have to take into account
more rectangles but it should be easily deducible from the picture of
the situation. - In your case, when you subtract the second integral from the first one, you can get pretty much everything
between $-bd$ and $bd$. Just consider functions $x^n$ and inverses
$sqrt[n]{x}$ on $[0,1]$.
answered Jan 27 at 12:45
Roman HricRoman Hric
32115
$endgroup$
$begingroup$
Do you it does not have a certain answer ?
$endgroup$
– Hussien Mohamed
Jan 27 at 20:18
$begingroup$
As I wrote in the answer, take e.g. $f(x) = x^n$ on $[0,1]$, then its inverse is $g(y) = sqrt[n]{y}$ on $[0,1]$, and we have $int_{0}^{1} x^n , dx - int_{0}^{1} sqrt[n]{y} , dy = 2/n -1$.
$endgroup$
– Roman Hric
Jan 28 at 17:14
add a comment |
$begingroup$
For brevity, let's consider a special case: $f: [0,b] to [0,d]$ increasing, continuous, onto. If you draw a picture then you realize that the first integral is the area of the rectangle $R:=[0,b] times [0,d]$ under the graph of $f$ and the second integral is the area of $R$ above the graph - that's because the graph of the inverse is symmetrical about the diagonal and the second integral is the area under this graph. But if you reflect it around the diagonal, you get the graph of the original $f$ and you count the area "to the left" from it.
- So if you add them together, not subtract, then you get the area of
$R$, i.e. $bd$. In general situation, you have to take into account
more rectangles but it should be easily deducible from the picture of
the situation. - In your case, when you subtract the second integral from the first one, you can get pretty much everything
between $-bd$ and $bd$. Just consider functions $x^n$ and inverses
$sqrt[n]{x}$ on $[0,1]$.
answered Jan 27 at 12:45
Roman HricRoman Hric
32115
$endgroup$
For brevity, let's consider a special case: $f: [0,b] to [0,d]$ increasing, continuous, onto. If you draw a picture then you realize that the first integral is the area of the rectangle $R:=[0,b] times [0,d]$ under the graph of $f$ and the second integral is the area of $R$ above the graph - that's because the graph of the inverse is symmetrical about the diagonal and the second integral is the area under this graph. But if you reflect it around the diagonal, you get the graph of the original $f$ and you count the area "to the left" from it.
- So if you add them together, not subtract, then you get the area of
$R$, i.e. $bd$. In general situation, you have to take into account
more rectangles but it should be easily deducible from the picture of
the situation. - In your case, when you subtract the second integral from the first one, you can get pretty much everything
between $-bd$ and $bd$. Just consider functions $x^n$ and inverses
$sqrt[n]{x}$ on $[0,1]$.
answered Jan 27 at 12:45
Roman HricRoman Hric
32115
answered Jan 27 at 12:45
Roman HricRoman Hric
32115
answered Jan 27 at 12:45
Roman HricRoman Hric
32115
answered Jan 27 at 12:45
Roman HricRoman Hric
32115
32115
$begingroup$
Do you it does not have a certain answer ?
$endgroup$
– Hussien Mohamed
Jan 27 at 20:18
$begingroup$
As I wrote in the answer, take e.g. $f(x) = x^n$ on $[0,1]$, then its inverse is $g(y) = sqrt[n]{y}$ on $[0,1]$, and we have $int_{0}^{1} x^n , dx - int_{0}^{1} sqrt[n]{y} , dy = 2/n -1$.
$endgroup$
– Roman Hric
Jan 28 at 17:14
add a comment |
$begingroup$
Do you it does not have a certain answer ?
$endgroup$
– Hussien Mohamed
Jan 27 at 20:18
$begingroup$
As I wrote in the answer, take e.g. $f(x) = x^n$ on $[0,1]$, then its inverse is $g(y) = sqrt[n]{y}$ on $[0,1]$, and we have $int_{0}^{1} x^n , dx - int_{0}^{1} sqrt[n]{y} , dy = 2/n -1$.
$endgroup$
– Roman Hric
Jan 28 at 17:14
$begingroup$
Do you it does not have a certain answer ?
$endgroup$
– Hussien Mohamed
Jan 27 at 20:18
$begingroup$
Do you it does not have a certain answer ?
$endgroup$
– Hussien Mohamed
Jan 27 at 20:18
$begingroup$
As I wrote in the answer, take e.g. $f(x) = x^n$ on $[0,1]$, then its inverse is $g(y) = sqrt[n]{y}$ on $[0,1]$, and we have $int_{0}^{1} x^n , dx - int_{0}^{1} sqrt[n]{y} , dy = 2/n -1$.
$endgroup$
– Roman Hric
Jan 28 at 17:14
$begingroup$
As I wrote in the answer, take e.g. $f(x) = x^n$ on $[0,1]$, then its inverse is $g(y) = sqrt[n]{y}$ on $[0,1]$, and we have $int_{0}^{1} x^n , dx - int_{0}^{1} sqrt[n]{y} , dy = 2/n -1$.
$endgroup$
– Roman Hric
Jan 28 at 17:14
add a comment |
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