# Integral resulting from the product of two functions/derivative functi

• I
• Saracen Rue
In summary: Though not explicitly stated, I believe the statement implies that the derivatives are well defined in the range of integration, just that the integrals have some sort of problem. (by "not valid" I think the OP means that the integral doesn't have a closed form)
Saracen Rue
TL;DR Summary
Define f(x) and g(x) so that the integral of f(x)*g'(x) is NOT valid but the integral of f'(x)*g(x) is.
Hey, sorry for the cluncky title. It was rathet difficult to summarise what I'm talking about here.

I want to know if it's possible to define ##f(x)## and ##g(x)## in such a way that ##∫f(x)g'(x)dx## has no indefinite solution while ##∫f'(x)g(x)dx## does have an indefinite solution.

Any help is greatly appreciated :)

Thank you all for your time.

Hi
Partial integral rule we write
$$\int^b_a fg'dx=[fg]^b_a - \int^b_a f'g dx$$
When left side integral and
$$[fg]^b_a =f(b)g(b)-f(a)g(a)$$
are definite, the right side integral is also definite.
Thus your require above fg be indefinite but I am afraid not.

Last edited:
Delta2
Saracen Rue said:
Summary: Define f(x) and g(x) so that the integral of f(x)*g'(x) is NOT valid but the integral of f'(x)*g(x) is.

Hey, sorry for the cluncky title. It was rathet difficult to summarise what I'm talking about here.

I want to know if it's possible to define ##f(x)## and ##g(x)## in such a way that ##∫f(x)g'(x)dx## has no indefinite solution while ##∫f'(x)g(x)dx## does have an indefinite solution.

Any help is greatly appreciated :)

Thank you all for your time.
What do you mean by "not valid"? Is the absolute value as ##g(x)## not valid, since ##g'(0)## doesn't exist?

fresh_42 said:
What do you mean by "not valid"? Is the absolute value as ##g(x)## not valid, since ##g'(0)## doesn't exist?
I meant "not valid" as to mean "have no results in terms of standard mathematical functions"

Riemann or Lebesgue? If it is Riemann, then ##g(x)=|x|## and a suitable ##f(x)## should do if ##0## is in the integration range.

fresh_42 said:
Riemann or Lebesgue? If it is Riemann, then ##g(x)=|x|## and a suitable ##f(x)## should do if ##0## is in the integration range.
Though not explicitly stated, I believe the statement implies that the derivatives are well defined in the range of integration, just that the integrals have some sort of problem. (by "not valid" I think the OP means that the integral doesn't have a closed form)

## 1. What is an integral resulting from the product of two functions?

An integral resulting from the product of two functions is a mathematical operation that involves finding the area under the curve of the product of two functions. It is denoted by ∫(f(x)g(x))dx and is used to solve various problems in calculus and physics.

## 2. How is an integral resulting from the product of two functions calculated?

An integral resulting from the product of two functions is calculated using the integration rules and techniques, such as substitution, integration by parts, and trigonometric substitution. It involves breaking down the product of two functions into simpler forms and then integrating each term separately.

## 3. What is the relationship between an integral resulting from the product of two functions and a derivative function?

The relationship between an integral resulting from the product of two functions and a derivative function is given by the fundamental theorem of calculus. It states that the integral of a derivative function is equal to the difference between the values of the original function at the upper and lower limits of integration.

## 4. What are the applications of integrals resulting from the product of two functions?

Integrals resulting from the product of two functions have various applications in mathematics, physics, and engineering. They are used to calculate the area under a curve, find the volume of a solid, determine the center of mass, and solve optimization problems.

## 5. Are there any limitations to using integrals resulting from the product of two functions?

One limitation of using integrals resulting from the product of two functions is that it can be challenging to find the exact value of the integral in some cases. This is because not all functions have an antiderivative that can be expressed in terms of elementary functions. In such cases, numerical methods can be used to approximate the integral.

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