Can an indefinite integral be expressed as a definite integral with variable bounds?












2














If I have a function $f(t)$, and an indefinite integral of this function, $g(x) = int f(t), dt$, is there any way I can express $g(x)$ as a definite integral whose bounds depend on $x$?



I thought I saw somewhere that I could express it as
$$g(x) = int_a^x f(t), dt$$
but I don't know what $a$ is. Is this representation correct? Or is there a better way to express $g$?










share|cite|improve this question
























  • The second representation is correct. The first one doesn't represent $g$ as a function of $x$.
    – user23793
    3 hours ago
















2














If I have a function $f(t)$, and an indefinite integral of this function, $g(x) = int f(t), dt$, is there any way I can express $g(x)$ as a definite integral whose bounds depend on $x$?



I thought I saw somewhere that I could express it as
$$g(x) = int_a^x f(t), dt$$
but I don't know what $a$ is. Is this representation correct? Or is there a better way to express $g$?










share|cite|improve this question
























  • The second representation is correct. The first one doesn't represent $g$ as a function of $x$.
    – user23793
    3 hours ago














2












2








2







If I have a function $f(t)$, and an indefinite integral of this function, $g(x) = int f(t), dt$, is there any way I can express $g(x)$ as a definite integral whose bounds depend on $x$?



I thought I saw somewhere that I could express it as
$$g(x) = int_a^x f(t), dt$$
but I don't know what $a$ is. Is this representation correct? Or is there a better way to express $g$?










share|cite|improve this question















If I have a function $f(t)$, and an indefinite integral of this function, $g(x) = int f(t), dt$, is there any way I can express $g(x)$ as a definite integral whose bounds depend on $x$?



I thought I saw somewhere that I could express it as
$$g(x) = int_a^x f(t), dt$$
but I don't know what $a$ is. Is this representation correct? Or is there a better way to express $g$?







calculus definite-integrals indefinite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Ethan Bolker

41.2k547108




41.2k547108










asked 3 hours ago









clabe45

1295




1295












  • The second representation is correct. The first one doesn't represent $g$ as a function of $x$.
    – user23793
    3 hours ago


















  • The second representation is correct. The first one doesn't represent $g$ as a function of $x$.
    – user23793
    3 hours ago
















The second representation is correct. The first one doesn't represent $g$ as a function of $x$.
– user23793
3 hours ago




The second representation is correct. The first one doesn't represent $g$ as a function of $x$.
– user23793
3 hours ago










2 Answers
2






active

oldest

votes


















5














No, not always.



In fact, the definite integral with variable bounds can be and is strictly more restrictive than the indefinite integral, in the most general case where both can be taken as sensibly existing to their fullest extent that we can say one defines the same kind of function as the other.



While it is true that if we have two functions $F_1$ and $F_2$ defined by definite integrals of the same function with different bounds, i.e.



$$F_1(x) := int_{a_1}^{x} f(t) dt mathrm{and} F_2(x) := int_{a_2}^{x} f(t) dt$$



with $a_1 ne a_2$, then we will have that $F_1 - F_2$ is a constant function, the range of possible such constants may be limited. A simple counterexample is to take $f(x) := sin(x)$. Now the integral from any lower bound $a$ to $x$ is just $int_a^x sin(t) dt = cos(a) - cos(x)$, but here $cos(a)$ - the "constant" of our integration - can only range in $[-1, 1]$ (and the difference of any two such "constants" only within $[-2, 2]$), but the constant in $int sin(x) dx = -cos(x) + C$ can be any real number whatsoever, even one far outside this range, say assign $C := mbox{Graham's number}$. Thus the integrals with variable bounds are not exhaustive of the full family of antiderivatives, which is what is represented by the indefinite integral, and if we are really pedantic should be truly written as the set it is: in "reality", not in the informalities of school calculus,



$$int f(x) dx = left{ F(x) + C : C in mathbb{R} right}$$



where $F'(x) = f(x)$. Actually, if we want to be really correct - and in fact this is important but only in cases where the definite integral technically doesn't make sense as evaluable between arbitrary pairs of points - we should take



$$int f(x) dx = left{ F(x) in mathbb{R}^mathbb{R} : mbox{$F'$ exists and $F'(x) = f(x)$} right}$$.






share|cite|improve this answer























  • +1 I thought about going to this level of sophistication in my much shorter answer but decided the OP needed just the central points I made. Now they can see both views.
    – Ethan Bolker
    3 hours ago










  • Very pedantic answer that is 100% valid and makes a distinction I had never thought about. +1 from me too. Heck, I'd give it a +5 if it was possible/allowed.
    – JonathanZ
    3 hours ago












  • I'm having a hard time understanding this, as I'm new to Calculus, but it's ok.
    – clabe45
    3 mins ago



















4














That fixed but arbitrary lower limit is essentially the "arbitrary constant" you use when finding antiderivatives, because
$$
g(x) = int_a^x f(t), dt =
int_b^x f(t), dt +
int_a^b f(t), dt .
$$

The second term in that sum is just a number.



The fundamental theorem of calculus says that both integrals with upper limit $x$ have derivative $f$ with respect to $x$.



The "indefinite integral" with no limits specified refers to the whole set of antiderivatives, not any particular one of them.






share|cite|improve this answer





















  • I'm almost following what you're saying, but not quite. I understand that both integrals with upper limit $x$ have derivative $f$ with respect to $x$, but how does that make the constant integral the arbitrary constant?
    – clabe45
    1 hour ago











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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5














No, not always.



In fact, the definite integral with variable bounds can be and is strictly more restrictive than the indefinite integral, in the most general case where both can be taken as sensibly existing to their fullest extent that we can say one defines the same kind of function as the other.



While it is true that if we have two functions $F_1$ and $F_2$ defined by definite integrals of the same function with different bounds, i.e.



$$F_1(x) := int_{a_1}^{x} f(t) dt mathrm{and} F_2(x) := int_{a_2}^{x} f(t) dt$$



with $a_1 ne a_2$, then we will have that $F_1 - F_2$ is a constant function, the range of possible such constants may be limited. A simple counterexample is to take $f(x) := sin(x)$. Now the integral from any lower bound $a$ to $x$ is just $int_a^x sin(t) dt = cos(a) - cos(x)$, but here $cos(a)$ - the "constant" of our integration - can only range in $[-1, 1]$ (and the difference of any two such "constants" only within $[-2, 2]$), but the constant in $int sin(x) dx = -cos(x) + C$ can be any real number whatsoever, even one far outside this range, say assign $C := mbox{Graham's number}$. Thus the integrals with variable bounds are not exhaustive of the full family of antiderivatives, which is what is represented by the indefinite integral, and if we are really pedantic should be truly written as the set it is: in "reality", not in the informalities of school calculus,



$$int f(x) dx = left{ F(x) + C : C in mathbb{R} right}$$



where $F'(x) = f(x)$. Actually, if we want to be really correct - and in fact this is important but only in cases where the definite integral technically doesn't make sense as evaluable between arbitrary pairs of points - we should take



$$int f(x) dx = left{ F(x) in mathbb{R}^mathbb{R} : mbox{$F'$ exists and $F'(x) = f(x)$} right}$$.






share|cite|improve this answer























  • +1 I thought about going to this level of sophistication in my much shorter answer but decided the OP needed just the central points I made. Now they can see both views.
    – Ethan Bolker
    3 hours ago










  • Very pedantic answer that is 100% valid and makes a distinction I had never thought about. +1 from me too. Heck, I'd give it a +5 if it was possible/allowed.
    – JonathanZ
    3 hours ago












  • I'm having a hard time understanding this, as I'm new to Calculus, but it's ok.
    – clabe45
    3 mins ago
















5














No, not always.



In fact, the definite integral with variable bounds can be and is strictly more restrictive than the indefinite integral, in the most general case where both can be taken as sensibly existing to their fullest extent that we can say one defines the same kind of function as the other.



While it is true that if we have two functions $F_1$ and $F_2$ defined by definite integrals of the same function with different bounds, i.e.



$$F_1(x) := int_{a_1}^{x} f(t) dt mathrm{and} F_2(x) := int_{a_2}^{x} f(t) dt$$



with $a_1 ne a_2$, then we will have that $F_1 - F_2$ is a constant function, the range of possible such constants may be limited. A simple counterexample is to take $f(x) := sin(x)$. Now the integral from any lower bound $a$ to $x$ is just $int_a^x sin(t) dt = cos(a) - cos(x)$, but here $cos(a)$ - the "constant" of our integration - can only range in $[-1, 1]$ (and the difference of any two such "constants" only within $[-2, 2]$), but the constant in $int sin(x) dx = -cos(x) + C$ can be any real number whatsoever, even one far outside this range, say assign $C := mbox{Graham's number}$. Thus the integrals with variable bounds are not exhaustive of the full family of antiderivatives, which is what is represented by the indefinite integral, and if we are really pedantic should be truly written as the set it is: in "reality", not in the informalities of school calculus,



$$int f(x) dx = left{ F(x) + C : C in mathbb{R} right}$$



where $F'(x) = f(x)$. Actually, if we want to be really correct - and in fact this is important but only in cases where the definite integral technically doesn't make sense as evaluable between arbitrary pairs of points - we should take



$$int f(x) dx = left{ F(x) in mathbb{R}^mathbb{R} : mbox{$F'$ exists and $F'(x) = f(x)$} right}$$.






share|cite|improve this answer























  • +1 I thought about going to this level of sophistication in my much shorter answer but decided the OP needed just the central points I made. Now they can see both views.
    – Ethan Bolker
    3 hours ago










  • Very pedantic answer that is 100% valid and makes a distinction I had never thought about. +1 from me too. Heck, I'd give it a +5 if it was possible/allowed.
    – JonathanZ
    3 hours ago












  • I'm having a hard time understanding this, as I'm new to Calculus, but it's ok.
    – clabe45
    3 mins ago














5












5








5






No, not always.



In fact, the definite integral with variable bounds can be and is strictly more restrictive than the indefinite integral, in the most general case where both can be taken as sensibly existing to their fullest extent that we can say one defines the same kind of function as the other.



While it is true that if we have two functions $F_1$ and $F_2$ defined by definite integrals of the same function with different bounds, i.e.



$$F_1(x) := int_{a_1}^{x} f(t) dt mathrm{and} F_2(x) := int_{a_2}^{x} f(t) dt$$



with $a_1 ne a_2$, then we will have that $F_1 - F_2$ is a constant function, the range of possible such constants may be limited. A simple counterexample is to take $f(x) := sin(x)$. Now the integral from any lower bound $a$ to $x$ is just $int_a^x sin(t) dt = cos(a) - cos(x)$, but here $cos(a)$ - the "constant" of our integration - can only range in $[-1, 1]$ (and the difference of any two such "constants" only within $[-2, 2]$), but the constant in $int sin(x) dx = -cos(x) + C$ can be any real number whatsoever, even one far outside this range, say assign $C := mbox{Graham's number}$. Thus the integrals with variable bounds are not exhaustive of the full family of antiderivatives, which is what is represented by the indefinite integral, and if we are really pedantic should be truly written as the set it is: in "reality", not in the informalities of school calculus,



$$int f(x) dx = left{ F(x) + C : C in mathbb{R} right}$$



where $F'(x) = f(x)$. Actually, if we want to be really correct - and in fact this is important but only in cases where the definite integral technically doesn't make sense as evaluable between arbitrary pairs of points - we should take



$$int f(x) dx = left{ F(x) in mathbb{R}^mathbb{R} : mbox{$F'$ exists and $F'(x) = f(x)$} right}$$.






share|cite|improve this answer














No, not always.



In fact, the definite integral with variable bounds can be and is strictly more restrictive than the indefinite integral, in the most general case where both can be taken as sensibly existing to their fullest extent that we can say one defines the same kind of function as the other.



While it is true that if we have two functions $F_1$ and $F_2$ defined by definite integrals of the same function with different bounds, i.e.



$$F_1(x) := int_{a_1}^{x} f(t) dt mathrm{and} F_2(x) := int_{a_2}^{x} f(t) dt$$



with $a_1 ne a_2$, then we will have that $F_1 - F_2$ is a constant function, the range of possible such constants may be limited. A simple counterexample is to take $f(x) := sin(x)$. Now the integral from any lower bound $a$ to $x$ is just $int_a^x sin(t) dt = cos(a) - cos(x)$, but here $cos(a)$ - the "constant" of our integration - can only range in $[-1, 1]$ (and the difference of any two such "constants" only within $[-2, 2]$), but the constant in $int sin(x) dx = -cos(x) + C$ can be any real number whatsoever, even one far outside this range, say assign $C := mbox{Graham's number}$. Thus the integrals with variable bounds are not exhaustive of the full family of antiderivatives, which is what is represented by the indefinite integral, and if we are really pedantic should be truly written as the set it is: in "reality", not in the informalities of school calculus,



$$int f(x) dx = left{ F(x) + C : C in mathbb{R} right}$$



where $F'(x) = f(x)$. Actually, if we want to be really correct - and in fact this is important but only in cases where the definite integral technically doesn't make sense as evaluable between arbitrary pairs of points - we should take



$$int f(x) dx = left{ F(x) in mathbb{R}^mathbb{R} : mbox{$F'$ exists and $F'(x) = f(x)$} right}$$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 3 hours ago

























answered 3 hours ago









The_Sympathizer

7,3802244




7,3802244












  • +1 I thought about going to this level of sophistication in my much shorter answer but decided the OP needed just the central points I made. Now they can see both views.
    – Ethan Bolker
    3 hours ago










  • Very pedantic answer that is 100% valid and makes a distinction I had never thought about. +1 from me too. Heck, I'd give it a +5 if it was possible/allowed.
    – JonathanZ
    3 hours ago












  • I'm having a hard time understanding this, as I'm new to Calculus, but it's ok.
    – clabe45
    3 mins ago


















  • +1 I thought about going to this level of sophistication in my much shorter answer but decided the OP needed just the central points I made. Now they can see both views.
    – Ethan Bolker
    3 hours ago










  • Very pedantic answer that is 100% valid and makes a distinction I had never thought about. +1 from me too. Heck, I'd give it a +5 if it was possible/allowed.
    – JonathanZ
    3 hours ago












  • I'm having a hard time understanding this, as I'm new to Calculus, but it's ok.
    – clabe45
    3 mins ago
















+1 I thought about going to this level of sophistication in my much shorter answer but decided the OP needed just the central points I made. Now they can see both views.
– Ethan Bolker
3 hours ago




+1 I thought about going to this level of sophistication in my much shorter answer but decided the OP needed just the central points I made. Now they can see both views.
– Ethan Bolker
3 hours ago












Very pedantic answer that is 100% valid and makes a distinction I had never thought about. +1 from me too. Heck, I'd give it a +5 if it was possible/allowed.
– JonathanZ
3 hours ago






Very pedantic answer that is 100% valid and makes a distinction I had never thought about. +1 from me too. Heck, I'd give it a +5 if it was possible/allowed.
– JonathanZ
3 hours ago














I'm having a hard time understanding this, as I'm new to Calculus, but it's ok.
– clabe45
3 mins ago




I'm having a hard time understanding this, as I'm new to Calculus, but it's ok.
– clabe45
3 mins ago











4














That fixed but arbitrary lower limit is essentially the "arbitrary constant" you use when finding antiderivatives, because
$$
g(x) = int_a^x f(t), dt =
int_b^x f(t), dt +
int_a^b f(t), dt .
$$

The second term in that sum is just a number.



The fundamental theorem of calculus says that both integrals with upper limit $x$ have derivative $f$ with respect to $x$.



The "indefinite integral" with no limits specified refers to the whole set of antiderivatives, not any particular one of them.






share|cite|improve this answer





















  • I'm almost following what you're saying, but not quite. I understand that both integrals with upper limit $x$ have derivative $f$ with respect to $x$, but how does that make the constant integral the arbitrary constant?
    – clabe45
    1 hour ago
















4














That fixed but arbitrary lower limit is essentially the "arbitrary constant" you use when finding antiderivatives, because
$$
g(x) = int_a^x f(t), dt =
int_b^x f(t), dt +
int_a^b f(t), dt .
$$

The second term in that sum is just a number.



The fundamental theorem of calculus says that both integrals with upper limit $x$ have derivative $f$ with respect to $x$.



The "indefinite integral" with no limits specified refers to the whole set of antiderivatives, not any particular one of them.






share|cite|improve this answer





















  • I'm almost following what you're saying, but not quite. I understand that both integrals with upper limit $x$ have derivative $f$ with respect to $x$, but how does that make the constant integral the arbitrary constant?
    – clabe45
    1 hour ago














4












4








4






That fixed but arbitrary lower limit is essentially the "arbitrary constant" you use when finding antiderivatives, because
$$
g(x) = int_a^x f(t), dt =
int_b^x f(t), dt +
int_a^b f(t), dt .
$$

The second term in that sum is just a number.



The fundamental theorem of calculus says that both integrals with upper limit $x$ have derivative $f$ with respect to $x$.



The "indefinite integral" with no limits specified refers to the whole set of antiderivatives, not any particular one of them.






share|cite|improve this answer












That fixed but arbitrary lower limit is essentially the "arbitrary constant" you use when finding antiderivatives, because
$$
g(x) = int_a^x f(t), dt =
int_b^x f(t), dt +
int_a^b f(t), dt .
$$

The second term in that sum is just a number.



The fundamental theorem of calculus says that both integrals with upper limit $x$ have derivative $f$ with respect to $x$.



The "indefinite integral" with no limits specified refers to the whole set of antiderivatives, not any particular one of them.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









Ethan Bolker

41.2k547108




41.2k547108












  • I'm almost following what you're saying, but not quite. I understand that both integrals with upper limit $x$ have derivative $f$ with respect to $x$, but how does that make the constant integral the arbitrary constant?
    – clabe45
    1 hour ago


















  • I'm almost following what you're saying, but not quite. I understand that both integrals with upper limit $x$ have derivative $f$ with respect to $x$, but how does that make the constant integral the arbitrary constant?
    – clabe45
    1 hour ago
















I'm almost following what you're saying, but not quite. I understand that both integrals with upper limit $x$ have derivative $f$ with respect to $x$, but how does that make the constant integral the arbitrary constant?
– clabe45
1 hour ago




I'm almost following what you're saying, but not quite. I understand that both integrals with upper limit $x$ have derivative $f$ with respect to $x$, but how does that make the constant integral the arbitrary constant?
– clabe45
1 hour ago


















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