Mixing
Fundamental Theorem of Calculus with Different Variables
$begingroup$
How can one explain that $$frac{d}{dx}left(int_0^x{cos(t^2+t)dt}right) = cos(x^2+x)$$
Without solving the integral?
I know it's related to the fundamental theorem of calculus, but here we have a derivative with respect to $x$, while the antiderivative is with respect to $t$.
Thank you.
calculus integration derivatives definite-integrals
edited Jan 13 at 16:43
KM101
6,0151524
asked Jan 13 at 16:34
NetanelNetanel
974
$endgroup$
add a comment |
$begingroup$
How can one explain that $$frac{d}{dx}left(int_0^x{cos(t^2+t)dt}right) = cos(x^2+x)$$
Without solving the integral?
I know it's related to the fundamental theorem of calculus, but here we have a derivative with respect to $x$, while the antiderivative is with respect to $t$.
Thank you.
calculus integration derivatives definite-integrals
edited Jan 13 at 16:43
KM101
6,0151524
asked Jan 13 at 16:34
NetanelNetanel
974
$endgroup$
1
$begingroup$
Related, possibly helpful: math.stackexchange.com/questions/3048884/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:35
add a comment |
$begingroup$
How can one explain that $$frac{d}{dx}left(int_0^x{cos(t^2+t)dt}right) = cos(x^2+x)$$
Without solving the integral?
I know it's related to the fundamental theorem of calculus, but here we have a derivative with respect to $x$, while the antiderivative is with respect to $t$.
Thank you.
calculus integration derivatives definite-integrals
edited Jan 13 at 16:43
KM101
6,0151524
asked Jan 13 at 16:34
NetanelNetanel
974
$endgroup$
How can one explain that $$frac{d}{dx}left(int_0^x{cos(t^2+t)dt}right) = cos(x^2+x)$$
Without solving the integral?
I know it's related to the fundamental theorem of calculus, but here we have a derivative with respect to $x$, while the antiderivative is with respect to $t$.
Thank you.
calculus integration derivatives definite-integrals
calculus integration derivatives definite-integrals
edited Jan 13 at 16:43
KM101
6,0151524
asked Jan 13 at 16:34
NetanelNetanel
974
edited Jan 13 at 16:43
KM101
6,0151524
asked Jan 13 at 16:34
NetanelNetanel
974
edited Jan 13 at 16:43
KM101
6,0151524
edited Jan 13 at 16:43
KM101
6,0151524
edited Jan 13 at 16:43
KM101
6,0151524
6,0151524
asked Jan 13 at 16:34
NetanelNetanel
974
asked Jan 13 at 16:34
NetanelNetanel
974
asked Jan 13 at 16:34
NetanelNetanel
974
974
1
$begingroup$
Related, possibly helpful: math.stackexchange.com/questions/3048884/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:35
add a comment |
1
$begingroup$
Related, possibly helpful: math.stackexchange.com/questions/3048884/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:35
1
1
$begingroup$
Related, possibly helpful: math.stackexchange.com/questions/3048884/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:35
$begingroup$
Related, possibly helpful: math.stackexchange.com/questions/3048884/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:35
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Say $f(t)=cos(t^2+t)$ and an antiderivative is $F(t)$. The integral in question is, by the fundamental theorem of calculus,
$$F(x)-F(0)$$
$F(0)$ is a constant and disappears upon differentiating with respect to $x$, whereas $F(x)$ becomes $f(x)$ once again. Thus, after differentiation we must have the RHS as $cos(x^2+x)$.
answered Jan 13 at 16:41
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
2
$begingroup$
An antiderivative $F$ exists because $f$ is continuous on the closed interval $[0, x]$. This proposition is sometimes referred to as the first part of the fundamental theorem of calculus, namely $x mapsto int_0^x f(t) mathrm{d}t$ is such an antiderivative.
$endgroup$
– ComFreek
Jan 13 at 17:12
add a comment |
$begingroup$
Perhaps you are mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC). First part of the theorem deals exactly with finding derivative of things of the form $int_{a} ^{x} f(t) , dt$.
More formally, let the function $f:[a, b] tomathbb {R} $ be Riemann integrable on $[a, b] $. The intent of the first part of FTC is to study the properties of a related function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ The function $F$ is defined by means of an integral and is not necessarily an anti-derivative of $f$.
FTC says that this function $F$ is continuous on $[a, b] $. And more importantly $F$ is differentiable at those points where $f$ is continuous and at such a point $cin[a, b] $ we have $F'(c) =f(c) $.
FTC does not say anything about $F'(c) $ when $f$ is discontinuous at $c$ and it may be possible that in such cases
$F'(c) $ does not exist- or it exists but does not equal $f(c) $
- or it may exist and be equal to $f(c) $
In any case one should observe that first part of FTC does not deal with anti-derivatives.
Now your function under the integral namely $cos(t^2+t)$ is continuous everywhere and hence the integral $int_{0}^{x}cos(t^2+t),dt$ is differentiable everywhere with derivative $cos(x^2+x)$. Thus the result in your question is an immediate consequence of the first part of FTC.
There is a second part of FTC which deals with anti-derivatives. Like the first part it begins with a function $f:[a, b] to mathbb {R} $ which is Riemann integrable on $[a, b] $ but its intent is to evaluate the integral $int_{a} ^{b} f(x) , dx$ in an easy manner. But to achieve this goal it makes an additional and strong assumption: it assumes that there is an anti-derivative $F$ of $f$ on $[a, b] $. In other words we assume the existence of a function $F:[a, b] tomathbb {R} $ such that $F'(x) =f(x) , forall xin[a, b] $ and then FTC says that the integral $int_{a} ^{b} f(x) , dx$ equals $F(b) - F(a) $.
answered Jan 13 at 17:27
Paramanand SinghParamanand Singh
50.1k556163
$endgroup$
add a comment |
$begingroup$
By the fundamental theorem of calculus, $int_a^b{f(t)dt}=F(b)-F(a)$, where $F'(x)=f(x)$. In this case:
$$int_0^x{cos(t^2+t)dt}=F(x)-F(0)$$
$$frac{d}{dx}(F(x)-F(0))=F'(x)=f(x)=cos(x^2+x)$$
$F(0)$ disappears after differentiation because it is a constant
answered Jan 13 at 16:48
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Say $f(t)=cos(t^2+t)$ and an antiderivative is $F(t)$. The integral in question is, by the fundamental theorem of calculus,
$$F(x)-F(0)$$
$F(0)$ is a constant and disappears upon differentiating with respect to $x$, whereas $F(x)$ becomes $f(x)$ once again. Thus, after differentiation we must have the RHS as $cos(x^2+x)$.
answered Jan 13 at 16:41
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
2
$begingroup$
An antiderivative $F$ exists because $f$ is continuous on the closed interval $[0, x]$. This proposition is sometimes referred to as the first part of the fundamental theorem of calculus, namely $x mapsto int_0^x f(t) mathrm{d}t$ is such an antiderivative.
$endgroup$
– ComFreek
Jan 13 at 17:12
add a comment |
$begingroup$
Say $f(t)=cos(t^2+t)$ and an antiderivative is $F(t)$. The integral in question is, by the fundamental theorem of calculus,
$$F(x)-F(0)$$
$F(0)$ is a constant and disappears upon differentiating with respect to $x$, whereas $F(x)$ becomes $f(x)$ once again. Thus, after differentiation we must have the RHS as $cos(x^2+x)$.
answered Jan 13 at 16:41
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
2
$begingroup$
An antiderivative $F$ exists because $f$ is continuous on the closed interval $[0, x]$. This proposition is sometimes referred to as the first part of the fundamental theorem of calculus, namely $x mapsto int_0^x f(t) mathrm{d}t$ is such an antiderivative.
$endgroup$
– ComFreek
Jan 13 at 17:12
add a comment |
$begingroup$
Say $f(t)=cos(t^2+t)$ and an antiderivative is $F(t)$. The integral in question is, by the fundamental theorem of calculus,
$$F(x)-F(0)$$
$F(0)$ is a constant and disappears upon differentiating with respect to $x$, whereas $F(x)$ becomes $f(x)$ once again. Thus, after differentiation we must have the RHS as $cos(x^2+x)$.
answered Jan 13 at 16:41
Parcly TaxelParcly Taxel
41.8k1372101
$endgroup$
Say $f(t)=cos(t^2+t)$ and an antiderivative is $F(t)$. The integral in question is, by the fundamental theorem of calculus,
$$F(x)-F(0)$$
$F(0)$ is a constant and disappears upon differentiating with respect to $x$, whereas $F(x)$ becomes $f(x)$ once again. Thus, after differentiation we must have the RHS as $cos(x^2+x)$.
answered Jan 13 at 16:41
Parcly TaxelParcly Taxel
41.8k1372101
answered Jan 13 at 16:41
Parcly TaxelParcly Taxel
41.8k1372101
answered Jan 13 at 16:41
Parcly TaxelParcly Taxel
41.8k1372101
answered Jan 13 at 16:41
Parcly TaxelParcly Taxel
41.8k1372101
41.8k1372101
2
$begingroup$
An antiderivative $F$ exists because $f$ is continuous on the closed interval $[0, x]$. This proposition is sometimes referred to as the first part of the fundamental theorem of calculus, namely $x mapsto int_0^x f(t) mathrm{d}t$ is such an antiderivative.
$endgroup$
– ComFreek
Jan 13 at 17:12
add a comment |
2
$begingroup$
An antiderivative $F$ exists because $f$ is continuous on the closed interval $[0, x]$. This proposition is sometimes referred to as the first part of the fundamental theorem of calculus, namely $x mapsto int_0^x f(t) mathrm{d}t$ is such an antiderivative.
$endgroup$
– ComFreek
Jan 13 at 17:12
2
2
$begingroup$
An antiderivative $F$ exists because $f$ is continuous on the closed interval $[0, x]$. This proposition is sometimes referred to as the first part of the fundamental theorem of calculus, namely $x mapsto int_0^x f(t) mathrm{d}t$ is such an antiderivative.
$endgroup$
– ComFreek
Jan 13 at 17:12
$begingroup$
An antiderivative $F$ exists because $f$ is continuous on the closed interval $[0, x]$. This proposition is sometimes referred to as the first part of the fundamental theorem of calculus, namely $x mapsto int_0^x f(t) mathrm{d}t$ is such an antiderivative.
$endgroup$
– ComFreek
Jan 13 at 17:12
add a comment |
$begingroup$
Perhaps you are mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC). First part of the theorem deals exactly with finding derivative of things of the form $int_{a} ^{x} f(t) , dt$.
More formally, let the function $f:[a, b] tomathbb {R} $ be Riemann integrable on $[a, b] $. The intent of the first part of FTC is to study the properties of a related function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ The function $F$ is defined by means of an integral and is not necessarily an anti-derivative of $f$.
FTC says that this function $F$ is continuous on $[a, b] $. And more importantly $F$ is differentiable at those points where $f$ is continuous and at such a point $cin[a, b] $ we have $F'(c) =f(c) $.
FTC does not say anything about $F'(c) $ when $f$ is discontinuous at $c$ and it may be possible that in such cases
$F'(c) $ does not exist- or it exists but does not equal $f(c) $
- or it may exist and be equal to $f(c) $
In any case one should observe that first part of FTC does not deal with anti-derivatives.
Now your function under the integral namely $cos(t^2+t)$ is continuous everywhere and hence the integral $int_{0}^{x}cos(t^2+t),dt$ is differentiable everywhere with derivative $cos(x^2+x)$. Thus the result in your question is an immediate consequence of the first part of FTC.
There is a second part of FTC which deals with anti-derivatives. Like the first part it begins with a function $f:[a, b] to mathbb {R} $ which is Riemann integrable on $[a, b] $ but its intent is to evaluate the integral $int_{a} ^{b} f(x) , dx$ in an easy manner. But to achieve this goal it makes an additional and strong assumption: it assumes that there is an anti-derivative $F$ of $f$ on $[a, b] $. In other words we assume the existence of a function $F:[a, b] tomathbb {R} $ such that $F'(x) =f(x) , forall xin[a, b] $ and then FTC says that the integral $int_{a} ^{b} f(x) , dx$ equals $F(b) - F(a) $.
answered Jan 13 at 17:27
Paramanand SinghParamanand Singh
50.1k556163
$endgroup$
add a comment |
$begingroup$
Perhaps you are mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC). First part of the theorem deals exactly with finding derivative of things of the form $int_{a} ^{x} f(t) , dt$.
More formally, let the function $f:[a, b] tomathbb {R} $ be Riemann integrable on $[a, b] $. The intent of the first part of FTC is to study the properties of a related function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ The function $F$ is defined by means of an integral and is not necessarily an anti-derivative of $f$.
FTC says that this function $F$ is continuous on $[a, b] $. And more importantly $F$ is differentiable at those points where $f$ is continuous and at such a point $cin[a, b] $ we have $F'(c) =f(c) $.
FTC does not say anything about $F'(c) $ when $f$ is discontinuous at $c$ and it may be possible that in such cases
$F'(c) $ does not exist- or it exists but does not equal $f(c) $
- or it may exist and be equal to $f(c) $
In any case one should observe that first part of FTC does not deal with anti-derivatives.
Now your function under the integral namely $cos(t^2+t)$ is continuous everywhere and hence the integral $int_{0}^{x}cos(t^2+t),dt$ is differentiable everywhere with derivative $cos(x^2+x)$. Thus the result in your question is an immediate consequence of the first part of FTC.
There is a second part of FTC which deals with anti-derivatives. Like the first part it begins with a function $f:[a, b] to mathbb {R} $ which is Riemann integrable on $[a, b] $ but its intent is to evaluate the integral $int_{a} ^{b} f(x) , dx$ in an easy manner. But to achieve this goal it makes an additional and strong assumption: it assumes that there is an anti-derivative $F$ of $f$ on $[a, b] $. In other words we assume the existence of a function $F:[a, b] tomathbb {R} $ such that $F'(x) =f(x) , forall xin[a, b] $ and then FTC says that the integral $int_{a} ^{b} f(x) , dx$ equals $F(b) - F(a) $.
answered Jan 13 at 17:27
Paramanand SinghParamanand Singh
50.1k556163
$endgroup$
add a comment |
$begingroup$
Perhaps you are mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC). First part of the theorem deals exactly with finding derivative of things of the form $int_{a} ^{x} f(t) , dt$.
More formally, let the function $f:[a, b] tomathbb {R} $ be Riemann integrable on $[a, b] $. The intent of the first part of FTC is to study the properties of a related function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ The function $F$ is defined by means of an integral and is not necessarily an anti-derivative of $f$.
FTC says that this function $F$ is continuous on $[a, b] $. And more importantly $F$ is differentiable at those points where $f$ is continuous and at such a point $cin[a, b] $ we have $F'(c) =f(c) $.
FTC does not say anything about $F'(c) $ when $f$ is discontinuous at $c$ and it may be possible that in such cases
$F'(c) $ does not exist- or it exists but does not equal $f(c) $
- or it may exist and be equal to $f(c) $
In any case one should observe that first part of FTC does not deal with anti-derivatives.
Now your function under the integral namely $cos(t^2+t)$ is continuous everywhere and hence the integral $int_{0}^{x}cos(t^2+t),dt$ is differentiable everywhere with derivative $cos(x^2+x)$. Thus the result in your question is an immediate consequence of the first part of FTC.
There is a second part of FTC which deals with anti-derivatives. Like the first part it begins with a function $f:[a, b] to mathbb {R} $ which is Riemann integrable on $[a, b] $ but its intent is to evaluate the integral $int_{a} ^{b} f(x) , dx$ in an easy manner. But to achieve this goal it makes an additional and strong assumption: it assumes that there is an anti-derivative $F$ of $f$ on $[a, b] $. In other words we assume the existence of a function $F:[a, b] tomathbb {R} $ such that $F'(x) =f(x) , forall xin[a, b] $ and then FTC says that the integral $int_{a} ^{b} f(x) , dx$ equals $F(b) - F(a) $.
answered Jan 13 at 17:27
Paramanand SinghParamanand Singh
50.1k556163
$endgroup$
Perhaps you are mixing two parts of the Fundamental Theorem of Calculus (henceforth referred to as FTC). First part of the theorem deals exactly with finding derivative of things of the form $int_{a} ^{x} f(t) , dt$.
More formally, let the function $f:[a, b] tomathbb {R} $ be Riemann integrable on $[a, b] $. The intent of the first part of FTC is to study the properties of a related function $F:[a, b] tomathbb {R} $ defined by $$F(x) =int_{a} ^{x} f(t) , dt$$ The function $F$ is defined by means of an integral and is not necessarily an anti-derivative of $f$.
FTC says that this function $F$ is continuous on $[a, b] $. And more importantly $F$ is differentiable at those points where $f$ is continuous and at such a point $cin[a, b] $ we have $F'(c) =f(c) $.
FTC does not say anything about $F'(c) $ when $f$ is discontinuous at $c$ and it may be possible that in such cases
$F'(c) $ does not exist- or it exists but does not equal $f(c) $
- or it may exist and be equal to $f(c) $
In any case one should observe that first part of FTC does not deal with anti-derivatives.
Now your function under the integral namely $cos(t^2+t)$ is continuous everywhere and hence the integral $int_{0}^{x}cos(t^2+t),dt$ is differentiable everywhere with derivative $cos(x^2+x)$. Thus the result in your question is an immediate consequence of the first part of FTC.
There is a second part of FTC which deals with anti-derivatives. Like the first part it begins with a function $f:[a, b] to mathbb {R} $ which is Riemann integrable on $[a, b] $ but its intent is to evaluate the integral $int_{a} ^{b} f(x) , dx$ in an easy manner. But to achieve this goal it makes an additional and strong assumption: it assumes that there is an anti-derivative $F$ of $f$ on $[a, b] $. In other words we assume the existence of a function $F:[a, b] tomathbb {R} $ such that $F'(x) =f(x) , forall xin[a, b] $ and then FTC says that the integral $int_{a} ^{b} f(x) , dx$ equals $F(b) - F(a) $.
answered Jan 13 at 17:27
Paramanand SinghParamanand Singh
50.1k556163
edited Jan 14 at 3:47
answered Jan 13 at 17:27
Paramanand SinghParamanand Singh
50.1k556163
answered Jan 13 at 17:27
Paramanand SinghParamanand Singh
50.1k556163
answered Jan 13 at 17:27
Paramanand SinghParamanand Singh
50.1k556163
50.1k556163
add a comment |
add a comment |
$begingroup$
By the fundamental theorem of calculus, $int_a^b{f(t)dt}=F(b)-F(a)$, where $F'(x)=f(x)$. In this case:
$$int_0^x{cos(t^2+t)dt}=F(x)-F(0)$$
$$frac{d}{dx}(F(x)-F(0))=F'(x)=f(x)=cos(x^2+x)$$
$F(0)$ disappears after differentiation because it is a constant
answered Jan 13 at 16:48
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
add a comment |
$begingroup$
By the fundamental theorem of calculus, $int_a^b{f(t)dt}=F(b)-F(a)$, where $F'(x)=f(x)$. In this case:
$$int_0^x{cos(t^2+t)dt}=F(x)-F(0)$$
$$frac{d}{dx}(F(x)-F(0))=F'(x)=f(x)=cos(x^2+x)$$
$F(0)$ disappears after differentiation because it is a constant
answered Jan 13 at 16:48
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
add a comment |
$begingroup$
By the fundamental theorem of calculus, $int_a^b{f(t)dt}=F(b)-F(a)$, where $F'(x)=f(x)$. In this case:
$$int_0^x{cos(t^2+t)dt}=F(x)-F(0)$$
$$frac{d}{dx}(F(x)-F(0))=F'(x)=f(x)=cos(x^2+x)$$
$F(0)$ disappears after differentiation because it is a constant
answered Jan 13 at 16:48
Lorenzo B.Lorenzo B.
1,8402520
$endgroup$
By the fundamental theorem of calculus, $int_a^b{f(t)dt}=F(b)-F(a)$, where $F'(x)=f(x)$. In this case:
$$int_0^x{cos(t^2+t)dt}=F(x)-F(0)$$
$$frac{d}{dx}(F(x)-F(0))=F'(x)=f(x)=cos(x^2+x)$$
$F(0)$ disappears after differentiation because it is a constant
answered Jan 13 at 16:48
Lorenzo B.Lorenzo B.
1,8402520
answered Jan 13 at 16:48
Lorenzo B.Lorenzo B.
1,8402520
answered Jan 13 at 16:48
Lorenzo B.Lorenzo B.
1,8402520
answered Jan 13 at 16:48
Lorenzo B.Lorenzo B.
1,8402520
1,8402520
add a comment |
add a comment |
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$begingroup$
Related, possibly helpful: math.stackexchange.com/questions/3048884/…
$endgroup$
– Ethan Bolker
Jan 13 at 18:35