- #1

- 39

- 5

## Summary:

- Derivative of definite integral

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

- I
- Thread starter Ahmed Mehedi
- Start date

- #1

- 39

- 5

## Summary:

- Derivative of definite integral

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

- #2

- #3

- 14,351

- 6,724

##F## is a real number. It is not a function of ##x##.Summary::Derivative of definite integral

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

- #4

- 39

- 5

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

- #5

Math_QED

Science Advisor

Homework Helper

2019 Award

- 1,693

- 719

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

- 39

- 5

What I actually wants to know here is as the following:##F## is a real number. It is not a function of ##x##.

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

Math_QED

Science Advisor

Homework Helper

2019 Award

- 1,693

- 719

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##.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?

You will get equality

- #8

- 14,351

- 6,724

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

- #9

- 39

- 5

Now I wants to know about the indefinite integral. But, I would be happier if you discuss both.I suspect there is a lot missing regarding the context of this question. Are these definite or indefinite integrals?

- #10

- 14,351

- 6,724

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

- 39

- 5

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.Can you not look at some examples yourself?

- #12

- 14,351

- 6,724

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

- 39

- 5

I will try expanding both of them using TS and check. Thanks for your suggestion!What about looking at Taylor series?

- #14

- 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

- 14,351

- 6,724

What about:

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

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

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

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

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

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

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

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

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

- #16

- 39

- 5

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.What about:

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

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

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

- #17

- 14,351

- 6,724

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

- 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

- 14,351

- 6,724

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

- 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

- 14,351

- 6,724

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

- 13

- 6

- #23

- 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

- 39

- 5

@TheDS1337 @Math_QED @etotheipiYou saidindefiniteintegral. Hence the constant of integration ##C## in my post.

- #25

Mark44

Mentor

- 33,964

- 5,622

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

- Last Post

- Replies
- 8

- Views
- 960

- Replies
- 8

- Views
- 2K

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 1K

- Replies
- 11

- Views
- 26K

- Replies
- 18

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 2K

- Replies
- 51

- Views
- 13K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 1K