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

**dirkie** · April 6th 2009, 04:22 PM

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

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

**Mush** · April 6th 2009, 05:01 PM

Originally Posted by dirkie

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

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

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

$g(x) = 0$

$a = b$

**dirkie** · April 6th 2009, 06:37 PM

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

$g(x) = 0$

$a = b$

Thank you very much.

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

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

**Mush** · April 6th 2009, 06:40 PM

Originally Posted by dirkie

Thank you very much.

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

$g(x) = 0$ 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'.

**TheEmptySet** · April 6th 2009, 06:42 PM

Originally Posted by dirkie

Thank you very much.

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

$g(x) = 0$ 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

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

**dirkie** · April 6th 2009, 06:43 PM

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.

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