# Derivative of a definite integral

• I
• Ahmed Mehedi
In summary: 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 rule.
Ahmed Mehedi
TL;DR Summary
Derivative of definite integral
If $$F(x)=\int_{a}^{b}f(x)dx$$ implies $$F'(x)=\int_{a}^{b}f'(x)dx$$?

Ahmed Mehedi said:
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##.

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

Also, ##F(x) = \int_a^b f(x) dx## makes not really sense. ##x## is both a variable and an integration dummy variable. Rather you should write ##F(x) = \int_a^b f(y)dy## and then ##F## is a constant function so ##F'(x) = 0##.

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.

Delta2 and Ahmed Mehedi
PeroK said:
##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?

Ahmed Mehedi said:
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 up to constant.

Ahmed Mehedi
Ahmed Mehedi said:
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?

PeroK said:
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.

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

PeroK said:
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.

Ahmed Mehedi said:
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.
What about looking at Taylor series?

PeroK said:
What about looking at Taylor series?

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

I intuitively look at the following steps. I would be happy if you point me out my mistakes if there is any:

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

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

Ahmed Mehedi said:
I intuitively look at the following steps. I would be happy if you point me out my mistakes if there is any:

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

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

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

PeroK said:

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

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.

PeroK
Ahmed Mehedi said:
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.
I assumed ##f, g## are continuous.

PeroK said:
I assumed ##f, g## are continuous.
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.

Ahmed Mehedi said:
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.
If ##f## is not continuous, how are you going to differentiate it?

Ahmed Mehedi said:
Summary:: Derivative of definite integral

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

PeroK said:
If ##f## is not continuous, how are you going to differentiate it?
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)##

Ahmed Mehedi said:
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)##
Ahmed Mehedi said:
Now I wants to know about the indefinite integral. But, I would be happier if you discuss both.
You said indefinite integral. Hence the constant of integration ##C## in my post.

You are confusing an integrand with a function, notice that after integration of f(x), the dependence of x is totaly lost and you have a real number, hence F'(x) = 0

Ahmed Mehedi
TheDS1337 said:
You are confusing an integrand with a function, notice that after integration of f(x), the dependence of x is totaly lost and you have a real number, hence F'(x) = 0

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:
In post #1 you wrote this:
Ahmed Mehedi said:
If ##F(x)=\int_{a}^{b}f(x)dx## implies ##F'(x)=\int_{a}^{b}f'(x)dx##?
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:
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 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

Delta2 and Ahmed Mehedi
Ahmed Mehedi said:
I intuitively look at the following steps. I would be happy if you point me out my mistakes if there is any:

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

Above steps appear to serve the purpose. But I guess I have been missing something as it seems so straight forward.
Mark44 said:
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:

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

I believe your 3 steps are correct and consistent with the definition of the indefinite integral as the anti-derivative.

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 for 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##.

Ahmed Mehedi
Delta2 said:
I believe your 3 steps are correct and consistent with the definition of the indefinite integral as the anti-derivative.

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 for 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!

Delta2 said:
I believe your 3 steps are correct and consistent with the definition of the indefinite integral as the anti-derivative.

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 for 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!

Delta2

## 1. What is the definition of the derivative of a definite integral?

The derivative of a definite integral is the rate of change of the integral with respect to its upper limit. It represents the slope of the tangent line to the integral curve at a specific point.

## 2. How is the derivative of a definite integral calculated?

The derivative of a definite integral can be calculated using the Fundamental Theorem of Calculus, which states that the derivative of a definite integral is equal to the integrand evaluated at the upper limit of integration.

## 3. What is the significance of the derivative of a definite integral in real-world applications?

The derivative of a definite integral is used in various fields of science and engineering to analyze rates of change and predict future behavior. It is particularly useful in physics, economics, and engineering.

## 4. Can the derivative of a definite integral be negative?

Yes, the derivative of a definite integral can be negative. This indicates that the integral is decreasing at that specific point.

## 5. Is the derivative of a definite integral always equal to the original integrand?

No, the derivative of a definite integral is not always equal to the original integrand. The Fundamental Theorem of Calculus only applies when the upper limit of integration is a constant. If the upper limit is a variable, the derivative will include an additional term involving the derivative of the upper limit.

• Calculus
Replies
31
Views
1K
• Calculus
Replies
6
Views
1K
• Calculus
Replies
20
Views
2K
• Calculus
Replies
1
Views
1K
• Calculus
Replies
16
Views
3K
• Calculus
Replies
4
Views
658
• Calculus
Replies
4
Views
1K
• Calculus
Replies
4
Views
553
• Calculus
Replies
8
Views
520
• Calculus
Replies
4
Views
517