Derivative of indefinite integral vs. definite
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So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$.
But if I have $frac{d}{dx} int f(x)dx$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?
And if I have the $int frac{d}{dx} f(x)dx$ - it is just the function as well, correct?
Thanks
integration definite-integrals indefinite-integrals
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up vote
2
down vote
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So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$.
But if I have $frac{d}{dx} int f(x)dx$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?
And if I have the $int frac{d}{dx} f(x)dx$ - it is just the function as well, correct?
Thanks
integration definite-integrals indefinite-integrals
1
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
– Ethan Bolker
6 hours ago
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
– user587192
4 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$.
But if I have $frac{d}{dx} int f(x)dx$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?
And if I have the $int frac{d}{dx} f(x)dx$ - it is just the function as well, correct?
Thanks
integration definite-integrals indefinite-integrals
So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$.
But if I have $frac{d}{dx} int f(x)dx$ - an indefinite integral -the derivative and the integral cancel each other out, and I just have the function: is that right?
And if I have the $int frac{d}{dx} f(x)dx$ - it is just the function as well, correct?
Thanks
integration definite-integrals indefinite-integrals
integration definite-integrals indefinite-integrals
edited 6 hours ago
Eevee Trainer
3,115224
3,115224
asked 6 hours ago
Josh White
543
543
1
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
– Ethan Bolker
6 hours ago
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
– user587192
4 hours ago
add a comment |
1
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
– Ethan Bolker
6 hours ago
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
– user587192
4 hours ago
1
1
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
– Ethan Bolker
6 hours ago
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
– Ethan Bolker
6 hours ago
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
– user587192
4 hours ago
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
– user587192
4 hours ago
add a comment |
2 Answers
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Not quite. The arbitrary constant from integration changes it up. Explicitly,
$$frac{d}{dx} int f(x)dx = f(x)$$
$$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$
An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.
Then
$$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$
but
$$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$
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0
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For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.
For the second question, you have to add the Constant of Integration to the function because of the rules of integration.
add a comment |
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2 Answers
2
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2 Answers
2
active
oldest
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active
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active
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votes
up vote
4
down vote
Not quite. The arbitrary constant from integration changes it up. Explicitly,
$$frac{d}{dx} int f(x)dx = f(x)$$
$$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$
An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.
Then
$$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$
but
$$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$
add a comment |
up vote
4
down vote
Not quite. The arbitrary constant from integration changes it up. Explicitly,
$$frac{d}{dx} int f(x)dx = f(x)$$
$$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$
An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.
Then
$$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$
but
$$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$
add a comment |
up vote
4
down vote
up vote
4
down vote
Not quite. The arbitrary constant from integration changes it up. Explicitly,
$$frac{d}{dx} int f(x)dx = f(x)$$
$$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$
An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.
Then
$$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$
but
$$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$
Not quite. The arbitrary constant from integration changes it up. Explicitly,
$$frac{d}{dx} int f(x)dx = f(x)$$
$$int left( frac{d}{dx} f(x) right) dx = f(x) + C$$
An example will be illustrative in finding the anomaly if it's not obvious. Let $f(x) = x^2$. We know $int f(x)dx = x^3/3 + C$, $f'(x) = 2x$, and, obviously, the derivative of a constant $C$ is $0$.
Then
$$frac{d}{dx} int f(x)dx = frac{d}{dx} int x^2dx = frac{d}{dx} left( frac{x^3}{3} + C right) = frac{3x^2}{3} + 0 = x^2 = f(x)$$
but
$$int left( frac{d}{dx} f(x) right) dx = int left( frac{d}{dx} x^2 right) dx = int 2xdx = frac{2x^2}{2} + C = x^2 + C = f(x) + C$$
answered 6 hours ago
Eevee Trainer
3,115224
3,115224
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add a comment |
up vote
0
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For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.
For the second question, you have to add the Constant of Integration to the function because of the rules of integration.
add a comment |
up vote
0
down vote
For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.
For the second question, you have to add the Constant of Integration to the function because of the rules of integration.
add a comment |
up vote
0
down vote
up vote
0
down vote
For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.
For the second question, you have to add the Constant of Integration to the function because of the rules of integration.
For the first question, I believe that it is 0 because the definite integral evaluates to a number, and the derivative of a number (constant) is 0. Not very sure on that one though.
For the second question, you have to add the Constant of Integration to the function because of the rules of integration.
answered 6 hours ago
Michael Wang
188
188
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1
You are trying to capture the subtleties of the fundamental theorem of calculus in a few pithy rules. You have it mostly right, but you should probably try to understand those rules in order to use them properly. en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
– Ethan Bolker
6 hours ago
"So I understand what to do if I have $frac{d}{dx} int_a^b f(x)dx$ for some function $f$" If you did not get $0$, then you were thinking in a wrong way.
– user587192
4 hours ago