# Thread: if \int a(x) b(x) dx = 0, is \int b(x) dx = 0 ?

1. ## if \int a(x) b(x) dx = 0, is \int b(x) dx = 0 ?

Assume $\displaystyle \int a(x) b(x) dx = 0$ and $\displaystyle a(x) \neq 0$.
Can this be simplified to $\displaystyle \int b(x) dx = 0$?

Or in another way, if
$\displaystyle \int f(x) g(x) dx = \int f(x) h(x) dx$,
then does this hold:
$\displaystyle \int g(x) dx = \int h(x) dx$

2. Originally Posted by dirkie
Assume $\displaystyle \int a(x) b(x) dx = 0$ and $\displaystyle a(x) \neq 0$.
Can this be simplified to $\displaystyle \int b(x) dx = 0$?

Or in another way, if
$\displaystyle \int f(x) g(x) dx = \int f(x) h(x) dx$,
then does this hold:
$\displaystyle \int g(x) dx = \int h(x) dx$
No, they can't be simplified to that, unless a(x) and f(x) are constants.

And in general, indefinite integrals aren't equal to zero as an arbitrary constant is always added to the result.

If the integral was definite, there are still a variety of reasons why it would be equal to zero:

If $\displaystyle \int_a^b f(x)g(x)dx = 0$, with $\displaystyle f(x) \neq 0$ then some possibilities are:

$\displaystyle g(x) = 0$

$\displaystyle a = b$

3. Originally Posted by Mush
No, they can't be simplified to that, unless a(x) and f(x) are constants.

And in general, indefinite integrals aren't equal to zero as an arbitrary constant is always added to the result.

If the integral was definite, there are still a variety of reasons why it would be equal to zero:

If $\displaystyle \int_a^b f(x)g(x)dx = 0$, with $\displaystyle f(x) \neq 0$ then some possibilities are:

$\displaystyle g(x) = 0$

$\displaystyle a = b$

Thank you very much.

If the integral was definite, i.e. $\displaystyle \int_a^b f(x)g(x)dx = 0$

$\displaystyle g(x) = 0$ is one solution, but not the only one, or is it?

4. Originally Posted by dirkie
Thank you very much.

If the integral was definite, i.e. $\displaystyle \int_a^b f(x)g(x)dx = 0$

$\displaystyle g(x) = 0$ is one solution, but not the only one, or is it?
No. Like I said, the integral will also be zero if the limits are equal a = b.

And it can also 'just happen'.

5. Originally Posted by dirkie
Thank you very much.

If the integral was definite, i.e. $\displaystyle \int_a^b f(x)g(x)dx = 0$

$\displaystyle g(x) = 0$ is one solution, but not the only one, or is it?
There are an infinite number of possiblilities.

Here is one class of examples

If the interval is symmetric about the origin (i.e. a=-b)then

$\displaystyle \int_{a}^{b}f(x)g(x)dx=0$ if f is an even function and g is an odd function

6. Originally Posted by Mush
No. Like I said, the integral will also be zero if the limits are equal a = b.

And it can also 'just happen'.
Thanks. I am expecting the 'just happen' scenario for the problem, and was making sure if it exists. So it does.