Derivative of a definite integral

[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]

@PeroK @dRic2 @etotheipi @Math_QED @Adesh

 

[PeroK]

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 $x$.

 

[Ahmed Mehedi]

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

Sorry. I understood after posting it.

 

[member 587159]

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.

 

[Ahmed Mehedi]

$F$ 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?

 

[member 587159]

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.

 

[PeroK]

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?

 

[Ahmed Mehedi]

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.

 

[PeroK]

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?

 

[Ahmed Mehedi]

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.

 

[PeroK]

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?

 

[Ahmed Mehedi]

What about looking at Taylor series?

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

 

[Ahmed Mehedi]

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.

 

[PeroK]

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.

What about:

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

 

[Ahmed Mehedi]

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.

 

[PeroK]

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.

 

[Ahmed Mehedi]

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.

 

[PeroK]

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?

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]

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 $f(x)\neq g(x)$

 

[PeroK]

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

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.

 

[TheDS1337]

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]

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.

 

[Ahmed Mehedi]

You said indefinite integral. Hence the constant of integration $C$ in my post.

@TheDS1337 @Math_QED @etotheipi

 

[Mark44]

In post #1 you wrote this:

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.

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

 

[Ahmed Mehedi]

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.

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:

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

 

[Delta2]

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]

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!

 

[Ahmed Mehedi]

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!