Antiderivative: does it represent any area?
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I was told that an antiderivative doesn't represent any area, it is just a family of functions, but today I noticed something quite interesting thing:
If we have $f(x)=2x$, then $F(x)=x^2+C$. The area under $f(x)$ from $0$ to $1$ is $1$, and $F(1)$ is one, the area under $f(x)$ from $0$ to $2$ is $4$, and $F(2)$ is $4$, and so on.
Then I started to think, that when $C=0$, $F(x)$ is the area under $f(x)$ from $0$ to $x$, and tried some others functions like $f(x)=x^2$, $x^3$, and so on.
Then I decided to try $f(x)=sin x$, and I failed in my guess. But then noticed, if I take $C=1$ it works! Finally, I noticed that it works in all cases when $F(0)=f(0)=0$.
So antiderivative is not only a bunch of functions but also some area?
How can I visualize it generally? How can I visualize $C$? Can this area be an endless area in both directions?
Now I study the fundamental theorem of calculus, and at first, I imagined $F(b)-F(a)$ as $(text{area from } -infty text{ or } 0 text{ to } b)-(text{ area from }-infty text{ or } 0 text{ to } a)$. Is it the right visualization?
I will be very grateful if someone could clarify my guesses. Thanks in advance
Upd: thank you all who helped me to solve my problem, i appreciate it so!
To one, who faced the same question: there`s a quite similar question, and some of answers are pretty meaningful, for example, this one:
https://math.stackexchange.com/a/2559329/595919
calculus integration area
asked Dec 18 '18 at 17:06
Max KnyazeffMax Knyazeff
134
$endgroup$
|
show 2 more comments
$begingroup$
I was told that an antiderivative doesn't represent any area, it is just a family of functions, but today I noticed something quite interesting thing:
If we have $f(x)=2x$, then $F(x)=x^2+C$. The area under $f(x)$ from $0$ to $1$ is $1$, and $F(1)$ is one, the area under $f(x)$ from $0$ to $2$ is $4$, and $F(2)$ is $4$, and so on.
Then I started to think, that when $C=0$, $F(x)$ is the area under $f(x)$ from $0$ to $x$, and tried some others functions like $f(x)=x^2$, $x^3$, and so on.
Then I decided to try $f(x)=sin x$, and I failed in my guess. But then noticed, if I take $C=1$ it works! Finally, I noticed that it works in all cases when $F(0)=f(0)=0$.
So antiderivative is not only a bunch of functions but also some area?
How can I visualize it generally? How can I visualize $C$? Can this area be an endless area in both directions?
Now I study the fundamental theorem of calculus, and at first, I imagined $F(b)-F(a)$ as $(text{area from } -infty text{ or } 0 text{ to } b)-(text{ area from }-infty text{ or } 0 text{ to } a)$. Is it the right visualization?
I will be very grateful if someone could clarify my guesses. Thanks in advance
Upd: thank you all who helped me to solve my problem, i appreciate it so!
To one, who faced the same question: there`s a quite similar question, and some of answers are pretty meaningful, for example, this one:
https://math.stackexchange.com/a/2559329/595919
calculus integration area
asked Dec 18 '18 at 17:06
Max KnyazeffMax Knyazeff
134
$endgroup$
$begingroup$
"I was told that an antiderivative doesn't represent any area": don't trust them.
$endgroup$
– Yves Daoust
Dec 18 '18 at 17:58
$begingroup$
So it does represent some area?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:00
$begingroup$
This is well-known.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:03
$begingroup$
So could u tell me please what area does it represent?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:05
$begingroup$
As usual, between the axis, curve and verticals.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:09
|
show 2 more comments
$begingroup$
I was told that an antiderivative doesn't represent any area, it is just a family of functions, but today I noticed something quite interesting thing:
If we have $f(x)=2x$, then $F(x)=x^2+C$. The area under $f(x)$ from $0$ to $1$ is $1$, and $F(1)$ is one, the area under $f(x)$ from $0$ to $2$ is $4$, and $F(2)$ is $4$, and so on.
Then I started to think, that when $C=0$, $F(x)$ is the area under $f(x)$ from $0$ to $x$, and tried some others functions like $f(x)=x^2$, $x^3$, and so on.
Then I decided to try $f(x)=sin x$, and I failed in my guess. But then noticed, if I take $C=1$ it works! Finally, I noticed that it works in all cases when $F(0)=f(0)=0$.
So antiderivative is not only a bunch of functions but also some area?
How can I visualize it generally? How can I visualize $C$? Can this area be an endless area in both directions?
Now I study the fundamental theorem of calculus, and at first, I imagined $F(b)-F(a)$ as $(text{area from } -infty text{ or } 0 text{ to } b)-(text{ area from }-infty text{ or } 0 text{ to } a)$. Is it the right visualization?
I will be very grateful if someone could clarify my guesses. Thanks in advance
Upd: thank you all who helped me to solve my problem, i appreciate it so!
To one, who faced the same question: there`s a quite similar question, and some of answers are pretty meaningful, for example, this one:
https://math.stackexchange.com/a/2559329/595919
calculus integration area
asked Dec 18 '18 at 17:06
Max KnyazeffMax Knyazeff
134
$endgroup$
I was told that an antiderivative doesn't represent any area, it is just a family of functions, but today I noticed something quite interesting thing:
If we have $f(x)=2x$, then $F(x)=x^2+C$. The area under $f(x)$ from $0$ to $1$ is $1$, and $F(1)$ is one, the area under $f(x)$ from $0$ to $2$ is $4$, and $F(2)$ is $4$, and so on.
Then I started to think, that when $C=0$, $F(x)$ is the area under $f(x)$ from $0$ to $x$, and tried some others functions like $f(x)=x^2$, $x^3$, and so on.
Then I decided to try $f(x)=sin x$, and I failed in my guess. But then noticed, if I take $C=1$ it works! Finally, I noticed that it works in all cases when $F(0)=f(0)=0$.
So antiderivative is not only a bunch of functions but also some area?
How can I visualize it generally? How can I visualize $C$? Can this area be an endless area in both directions?
Now I study the fundamental theorem of calculus, and at first, I imagined $F(b)-F(a)$ as $(text{area from } -infty text{ or } 0 text{ to } b)-(text{ area from }-infty text{ or } 0 text{ to } a)$. Is it the right visualization?
I will be very grateful if someone could clarify my guesses. Thanks in advance
Upd: thank you all who helped me to solve my problem, i appreciate it so!
To one, who faced the same question: there`s a quite similar question, and some of answers are pretty meaningful, for example, this one:
https://math.stackexchange.com/a/2559329/595919
calculus integration area
calculus integration area
asked Dec 18 '18 at 17:06
Max KnyazeffMax Knyazeff
134
asked Dec 18 '18 at 17:06
Max KnyazeffMax Knyazeff
134
edited Dec 19 '18 at 15:30
Max Knyazeff
asked Dec 18 '18 at 17:06
Max KnyazeffMax Knyazeff
134
asked Dec 18 '18 at 17:06
Max KnyazeffMax Knyazeff
134
asked Dec 18 '18 at 17:06
Max KnyazeffMax Knyazeff
134
134
$begingroup$
"I was told that an antiderivative doesn't represent any area": don't trust them.
$endgroup$
– Yves Daoust
Dec 18 '18 at 17:58
$begingroup$
So it does represent some area?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:00
$begingroup$
This is well-known.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:03
$begingroup$
So could u tell me please what area does it represent?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:05
$begingroup$
As usual, between the axis, curve and verticals.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:09
|
show 2 more comments
$begingroup$
"I was told that an antiderivative doesn't represent any area": don't trust them.
$endgroup$
– Yves Daoust
Dec 18 '18 at 17:58
$begingroup$
So it does represent some area?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:00
$begingroup$
This is well-known.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:03
$begingroup$
So could u tell me please what area does it represent?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:05
$begingroup$
As usual, between the axis, curve and verticals.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:09
$begingroup$
"I was told that an antiderivative doesn't represent any area": don't trust them.
$endgroup$
– Yves Daoust
Dec 18 '18 at 17:58
$begingroup$
"I was told that an antiderivative doesn't represent any area": don't trust them.
$endgroup$
– Yves Daoust
Dec 18 '18 at 17:58
$begingroup$
So it does represent some area?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:00
$begingroup$
So it does represent some area?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:00
$begingroup$
This is well-known.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:03
$begingroup$
This is well-known.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:03
$begingroup$
So could u tell me please what area does it represent?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:05
$begingroup$
So could u tell me please what area does it represent?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:05
$begingroup$
As usual, between the axis, curve and verticals.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:09
$begingroup$
As usual, between the axis, curve and verticals.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:09
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Let $fcolon Bbb RtoBbb R$ be a function that has an anti-derivative $F$. Let $a<b$ be real numbers, then you can use $F$ to calculate the (oriented) area between the vertical lines $x=a$ and $x=b$ that sits between the $x$-axis and the graph of $f$:
More precisely we have
$$
color{blue}{text{blue area}} - color{orange}{text{orange area}} = int_a^b f(x),mathrm dx = F(b) - F(a).
$$
Note that the choice of constant in the anti-derivative doesn't matter here, since it cancels when calculating the difference $F(b)-F(a)$.
Letting $a=0$ and choosing the constant such that $F(0)=0$ you get that $F(a)$ is the area under the curve between $x=0$ and $x=a$, as you figured out.
answered Dec 19 '18 at 15:45
ChristophChristoph
12.5k1642
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add a comment |
$begingroup$
Firstly, you can visualize an anti-derivative as a curve itself, whose derivative yields the function with which you are concerned. This new curve can be translated up and down without affecting the slope, therefore you can visualize C as an ambiguous translation of that curve up and down.
As for your understanding of the Fundamental Theorem of Calculus, I would not interpret it as two areas, but rather as the area between a and b itself, which can be found using your anti-derivative function.
I hope that helps.
answered Dec 18 '18 at 17:49
RJJBRJJB
132
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let $fcolon Bbb RtoBbb R$ be a function that has an anti-derivative $F$. Let $a<b$ be real numbers, then you can use $F$ to calculate the (oriented) area between the vertical lines $x=a$ and $x=b$ that sits between the $x$-axis and the graph of $f$:
More precisely we have
$$
color{blue}{text{blue area}} - color{orange}{text{orange area}} = int_a^b f(x),mathrm dx = F(b) - F(a).
$$
Note that the choice of constant in the anti-derivative doesn't matter here, since it cancels when calculating the difference $F(b)-F(a)$.
Letting $a=0$ and choosing the constant such that $F(0)=0$ you get that $F(a)$ is the area under the curve between $x=0$ and $x=a$, as you figured out.
answered Dec 19 '18 at 15:45
ChristophChristoph
12.5k1642
$endgroup$
add a comment |
$begingroup$
Let $fcolon Bbb RtoBbb R$ be a function that has an anti-derivative $F$. Let $a<b$ be real numbers, then you can use $F$ to calculate the (oriented) area between the vertical lines $x=a$ and $x=b$ that sits between the $x$-axis and the graph of $f$:
More precisely we have
$$
color{blue}{text{blue area}} - color{orange}{text{orange area}} = int_a^b f(x),mathrm dx = F(b) - F(a).
$$
Note that the choice of constant in the anti-derivative doesn't matter here, since it cancels when calculating the difference $F(b)-F(a)$.
Letting $a=0$ and choosing the constant such that $F(0)=0$ you get that $F(a)$ is the area under the curve between $x=0$ and $x=a$, as you figured out.
answered Dec 19 '18 at 15:45
ChristophChristoph
12.5k1642
$endgroup$
add a comment |
$begingroup$
Let $fcolon Bbb RtoBbb R$ be a function that has an anti-derivative $F$. Let $a<b$ be real numbers, then you can use $F$ to calculate the (oriented) area between the vertical lines $x=a$ and $x=b$ that sits between the $x$-axis and the graph of $f$:
More precisely we have
$$
color{blue}{text{blue area}} - color{orange}{text{orange area}} = int_a^b f(x),mathrm dx = F(b) - F(a).
$$
Note that the choice of constant in the anti-derivative doesn't matter here, since it cancels when calculating the difference $F(b)-F(a)$.
Letting $a=0$ and choosing the constant such that $F(0)=0$ you get that $F(a)$ is the area under the curve between $x=0$ and $x=a$, as you figured out.
answered Dec 19 '18 at 15:45
ChristophChristoph
12.5k1642
$endgroup$
Let $fcolon Bbb RtoBbb R$ be a function that has an anti-derivative $F$. Let $a<b$ be real numbers, then you can use $F$ to calculate the (oriented) area between the vertical lines $x=a$ and $x=b$ that sits between the $x$-axis and the graph of $f$:
More precisely we have
$$
color{blue}{text{blue area}} - color{orange}{text{orange area}} = int_a^b f(x),mathrm dx = F(b) - F(a).
$$
Note that the choice of constant in the anti-derivative doesn't matter here, since it cancels when calculating the difference $F(b)-F(a)$.
Letting $a=0$ and choosing the constant such that $F(0)=0$ you get that $F(a)$ is the area under the curve between $x=0$ and $x=a$, as you figured out.
answered Dec 19 '18 at 15:45
ChristophChristoph
12.5k1642
answered Dec 19 '18 at 15:45
ChristophChristoph
12.5k1642
answered Dec 19 '18 at 15:45
ChristophChristoph
12.5k1642
answered Dec 19 '18 at 15:45
ChristophChristoph
12.5k1642
12.5k1642
add a comment |
add a comment |
$begingroup$
Firstly, you can visualize an anti-derivative as a curve itself, whose derivative yields the function with which you are concerned. This new curve can be translated up and down without affecting the slope, therefore you can visualize C as an ambiguous translation of that curve up and down.
As for your understanding of the Fundamental Theorem of Calculus, I would not interpret it as two areas, but rather as the area between a and b itself, which can be found using your anti-derivative function.
I hope that helps.
answered Dec 18 '18 at 17:49
RJJBRJJB
132
$endgroup$
add a comment |
$begingroup$
Firstly, you can visualize an anti-derivative as a curve itself, whose derivative yields the function with which you are concerned. This new curve can be translated up and down without affecting the slope, therefore you can visualize C as an ambiguous translation of that curve up and down.
As for your understanding of the Fundamental Theorem of Calculus, I would not interpret it as two areas, but rather as the area between a and b itself, which can be found using your anti-derivative function.
I hope that helps.
answered Dec 18 '18 at 17:49
RJJBRJJB
132
$endgroup$
add a comment |
$begingroup$
Firstly, you can visualize an anti-derivative as a curve itself, whose derivative yields the function with which you are concerned. This new curve can be translated up and down without affecting the slope, therefore you can visualize C as an ambiguous translation of that curve up and down.
As for your understanding of the Fundamental Theorem of Calculus, I would not interpret it as two areas, but rather as the area between a and b itself, which can be found using your anti-derivative function.
I hope that helps.
answered Dec 18 '18 at 17:49
RJJBRJJB
132
$endgroup$
Firstly, you can visualize an anti-derivative as a curve itself, whose derivative yields the function with which you are concerned. This new curve can be translated up and down without affecting the slope, therefore you can visualize C as an ambiguous translation of that curve up and down.
As for your understanding of the Fundamental Theorem of Calculus, I would not interpret it as two areas, but rather as the area between a and b itself, which can be found using your anti-derivative function.
I hope that helps.
answered Dec 18 '18 at 17:49
RJJBRJJB
132
answered Dec 18 '18 at 17:49
RJJBRJJB
132
answered Dec 18 '18 at 17:49
RJJBRJJB
132
answered Dec 18 '18 at 17:49
RJJBRJJB
132
132
add a comment |
add a comment |
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$begingroup$
"I was told that an antiderivative doesn't represent any area": don't trust them.
$endgroup$
– Yves Daoust
Dec 18 '18 at 17:58
$begingroup$
So it does represent some area?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:00
$begingroup$
This is well-known.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:03
$begingroup$
So could u tell me please what area does it represent?
$endgroup$
– Max Knyazeff
Dec 18 '18 at 18:05
$begingroup$
As usual, between the axis, curve and verticals.
$endgroup$
– Yves Daoust
Dec 18 '18 at 18:09