- #1

Ahmed Mehedi

- 39

- 5

- TL;DR Summary
- Derivative of definite integral

If $$F(x)=\int_{a}^{b}f(x)dx$$ implies $$F'(x)=\int_{a}^{b}f'(x)dx$$?

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- Thread starter Ahmed Mehedi
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- #1

Ahmed Mehedi

- 39

- 5

- TL;DR Summary
- Derivative of definite integral

If $$F(x)=\int_{a}^{b}f(x)dx$$ implies $$F'(x)=\int_{a}^{b}f'(x)dx$$?

- #2

Ahmed Mehedi

- 39

- 5

- #3

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Summary::Derivative of definite integral

If $$F(x)=\int_{a}^{b}f(x)dx$$

##F## is a real number. It is not a function of ##x##.

- #4

Ahmed Mehedi

- 39

- 5

Sorry. I understood after posting it.##F## is a real number. It is not a function of ##x##.

- #5

As a side note, if you define ##F(x) = \int_a^x f(y)dy##, then ##F'(x) = f(x)##. This is the fundamental theorem of calculus.

- #6

Ahmed Mehedi

- 39

- 5

##F## is a real number. It is not a function of ##x##.

What I actually wants to know here is as the following:

If ##f## and ##g## are differentiable wrt ##x## and if $$\int g(x)dx=\int f(x)dx$$ can we write $$\int g'(x)dx=\int f'(x)dx$$? In short, can we pass the differentiation operator inside the integration operator without sacrifising the equality?

- #7

What I actually wants to know here is as the following:

If ##f## and ##g## are differentiable wrt ##x## and if $$\int g(x)dx=\int f(x)dx$$ can we write $$\int g'(x)dx=\int f'(x)dx$$? In short, can we pass the differentiation operator inside the integration operator without sacrifising the equality?

Let me first ask you a question: how do you define ##\int f(x)dx##? Formally, this is the set ##\{F: F' = f\}##. Assuming we work on an interval, we can rewrite this as ##\{F+c: c \in \mathbb{R}\}## where ##F' = f##.

You will get equality

- #8

- 24,026

- 15,718

What I actually wants to know here is as the following:

If ##f## and ##g## are differentiable wrt ##x## and if $$\int g(x)dx=\int f(x)dx$$ can we write $$\int g'(x)dx=\int f'(x)dx$$? In short, can we pass the differentiation operator inside the integration operator without sacrifising the equality?

I suspect there is a lot missing regarding the context of this question. Are these definite or indefinite integrals?

- #9

Ahmed Mehedi

- 39

- 5

I suspect there is a lot missing regarding the context of this question. Are these definite or indefinite integrals?

Now I wants to know about the indefinite integral. But, I would be happier if you discuss both.

- #10

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Can you not look at some examples yourself?Now I wants to know about the indefinite integral. But, I would be happier if you discuss both.

- #11

Ahmed Mehedi

- 39

- 5

Can you not look at some examples yourself?

I guess examples may reject or accept the claim. But they don't comprise a proof of the claim. I want to know whether it can be used as an identity irrespective of any special circumstances.

- #12

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What about looking at Taylor series?I guess examples may reject or accept the claim. But they don't comprise a proof of the claim. I want to know whether it can be used as an identity irrespective of any special circumstances.

- #13

Ahmed Mehedi

- 39

- 5

What about looking at Taylor series?

I will try expanding both of them using TS and check. Thanks for your suggestion!

- #14

Ahmed Mehedi

- 39

- 5

$$\int f(x)dx=\int g(x)dx$$

$$\frac{d}{dx}\int f(x)dx=\frac{d}{dx}\int g(x)dx$$

$$f(x)=g(x)$$

$$f'(x)=g'(x)$$

$$\int f'(x)dx=\int g'(x)dx$$

Above steps appear to serve the purpose. But I guess I have been missing something as it seems so straight forward.

- #15

- 24,026

- 15,718

$$\int f(x)dx=\int g(x)dx$$

$$\frac{d}{dx}\int f(x)dx=\frac{d}{dx}\int g(x)dx$$

$$f(x)=g(x)$$

$$f'(x)=g'(x)$$

$$\int f'(x)dx=\int g'(x)dx$$

Above steps appear to serve the purpose. But I guess I have been missing something as it seems so straight forward.

What about:

$$\int f(x)dx=\int g(x)dx + C$$

$$\int (f(x) - g(x))dx= C$$

$$ f(x) = g(x)$$

- #16

Ahmed Mehedi

- 39

- 5

What about:

$$\int f(x)dx=\int g(x)dx + C$$

$$\int (f(x) - g(x))dx= C$$

$$ f(x) = g(x)$$

This will also serve the purpose I guess. But, both of the approaches have one serious flaw. In the first line we are assuming equality between two integrals. While in the third line we are deriving the equality between the two integrands of the 1st line. That is we are claiming that equality of the two integrals confers the equality between the two integrands which can not be true as a general case.

- #17

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I assumed ##f, g## are continuous.This will also serve the purpose I guess. But, both of the approaches have one serious flaw. In the first line we are assuming equality between two integrals. While in the third line we are deriving the equality between the two integrands of the 1st line. That is we are claiming that equality of the two integrals confers the equality between the two integrands which can not be true as a general case.

- #18

Ahmed Mehedi

- 39

- 5

I guess that does not help much. g(t) can be any time-shifted version of f(t). Area under them may be equal while functions themselves are not.I assumed ##f, g## are continuous.

- #19

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If ##f## is not continuous, how are you going to differentiate it?I guess that does not help much. g(t) can be any time-shifted version of f(t). Area under them may be equal while functions themselves are not.

Summary::Derivative of definite integral

If $$F(x)=\int_{a}^{b}f(x)dx$$ implies $$F'(x)=\int_{a}^{b}f'(x)dx$$?

- #20

Ahmed Mehedi

- 39

- 5

Assume ##f(x)=x^2## and ##g(x)=(x-a)^2##. Both are continuous. Area under them are equal. Yet ##f(x)\neq g(x)##If ##f## is not continuous, how are you going to differentiate it?

- #21

- 24,026

- 15,718

Assume ##f(x)=x^2## and ##g(x)=(x-a)^2##. Both are continuous. Area under them are equal. Yet ##f(x)\neq g(x)##

You saidNow I wants to know about the indefinite integral. But, I would be happier if you discuss both.

- #22

TheDS1337

- 13

- 6

- #23

Ahmed Mehedi

- 39

- 5

Yes I got it. But the thing I really wanted to know is in the comments section. I am tagging you there.

Last edited by a moderator:

- #24

Ahmed Mehedi

- 39

- 5

You saidindefiniteintegral. Hence the constant of integration ##C## in my post.

@TheDS1337 @Math_QED @etotheipi

- #25

Mark44

Mentor

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As has already been pointed out, this is false. Here F(x) is a constant equal to the value of the definite integral on the right side of that equation. Hence, F'(x) = 0.Ahmed Mehedi said:If ##F(x)=\int_{a}^{b}f(x)dx## implies ##F'(x)=\int_{a}^{b}f'(x)dx##?

If you're talking about the area under these curves, you're necessarily talking about definite integrals. For the areas to be equal, the limits of integration would have to be different.Assume ##f(x)=x^2## and ##g(x)=(x-a)^2##. Both are continuous. Area under them are equal. Yet ##f(x)\neq g(x)##

For example ##\int_0^1 x^2 dx \ne \int_0^1 (x - a)^2 dx##, but ##\int_0^1 x^2 dx = \int_a^{1 + a}1 (x - a)^2 dx##.

You seem to be unclear on the differences between an indefinite integral, a definite integral, and an integral that represents a function.

Indefinite integral: ##\int f(x) dx## -- represents a function

Definite integral: ##\int_a^b f(x) dx## -- represesnts a number

Definite integral as a function: ##\int_a^t f(x)dx## -- represents a function of t

- #26

Ahmed Mehedi

- 39

- 5

$$\int f(x)dx=\int g(x)dx$$

$$\frac{d}{dx}\int f(x)dx=\frac{d}{dx}\int g(x)dx$$

$$f(x)=g(x)$$

$$f'(x)=g'(x)$$

$$\int f'(x)dx=\int g'(x)dx$$

Above steps appear to serve the purpose. But I guess I have been missing something as it seems so straight forward.

In post #1 you wrote this:

As has already been pointed out, this is false. Here F(x) is a constant equal to the value of the definite integral on the right side of that equation. Hence, F'(x) = 0.

If you're talking about the area under these curves, you're necessarily talking about definite integrals. For the areas to be equal, the limits of integration would have to be different.

For example ##\int_0^1 x^2 dx \ne \int_0^1 (x - a)^2 dx##, but ##\int_0^1 x^2 dx = \int_a^{1 + a}1 (x - a)^2 dx##.

You seem to be unclear on the differences between an indefinite integral, a definite integral, and an integral that represents a function.

Indefinite integral: ##\int f(x) dx## -- represents a function

Definite integral: ##\int_a^b f(x) dx## -- represesnts a number

Definite integral as a function: ##\int_a^t f(x)dx## -- represents a function of t

Thank you very much for your response! I have one last question in this regard. Are the following three steps correct in the context of indefinite integral:

$$\int f(x)dx=\int g(x)dx$$

$$\frac{d}{dx}\int f(x)dx=\frac{d}{dx}\int g(x)dx$$

$$f(x)=g(x)$$

- #27

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However if you going to attempt to do similar steps with definite integrals for example to start from

$$\int_a^b f(x)dx=\int_a^b g(x)dx \text { (1)}$$

then if you demand that the above equality holds

if (1) holds only for specific## a,b ##then you simply cannot infer that ##f(x)=g(x)## and taking the derivative with respect to x of (1) you ll just end up with ##0=0##.

- #28

Ahmed Mehedi

- 39

- 5

However if you going to attempt to do similar steps with definite integrals for example to start from

$$\int_a^b f(x)dx=\int_a^b g(x)dx \text { (1)}$$

then if you demand that the above equality holdsfor any##a,b## then you also can conclude that ##f(x)=g(x)##. We use an argument like this in classical electromagnetism to switch from the integral form of Maxwell's equations to the differential form of them.

if (1) holds only for specific## a,b ##then you simply cannot infer that ##f(x)=g(x)## and taking the derivative with respect to x of (1) you ll just end up with ##0=0##.

Thank you very much!

- #29

Ahmed Mehedi

- 39

- 5

However if you going to attempt to do similar steps with definite integrals for example to start from

$$\int_a^b f(x)dx=\int_a^b g(x)dx \text { (1)}$$

then if you demand that the above equality holdsfor any##a,b## then you also can conclude that ##f(x)=g(x)##. We use an argument like this in classical electromagnetism to switch from the integral form of Maxwell's equations to the differential form of them.

if (1) holds only for specific## a,b ##then you simply cannot infer that ##f(x)=g(x)## and taking the derivative with respect to x of (1) you ll just end up with ##0=0##.

Thank you very much for your revealing answers! Now, I clearly get the point!

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