Question

Let `f:[a,b]->RR^+uu{0}` where `int_0^bf(x)dx=0` then `f(x)=0` on `[a,b]` this sentence true or false ?

Answer 1

True.

Explanation:

Let `f(x)` be defined and continuous on `[a,b]` assuming only non negative values and:

`int_a^b f(x) dx = 0`

Consider now any point `xi in (a,b)`. Based on the additivity of the integral:

`(1) int_a^b f(x)dx = int_a^xi f(x)dx + int_xi^b f(x)dx`

Consider now the sum, that is necessarily positive:

`0 <= abs( int_a^xi f(x)dx) + abs(int_xi^b f(x)dx)`

Using the modulus inequality of integrals:

`abs( int_a^xi f(x)dx) + abs(int_xi^b f(x)dx) <= int_a^xi abs(f(x))dx+int_xi^b abs(f(x))dx`

As `f(x)` is non negative, `abs(f(x)) = f(x)`, so:

`abs( int_a^xi f(x)dx) + abs(int_xi^b f(x)dx) <= int_a^xi f(x)dx+int_xi^b f(x)dx`

that is:

`abs( int_a^xi f(x)dx) + abs(int_xi^b f(x)dx) <= int_a^b f(x)dx = 0`

So we have:

`0 <= abs( int_a^xi f(x)dx) + abs(int_xi^b f(x)dx) <= 0`

or:

`abs( int_a^xi f(x)dx) + abs(int_xi^b f(x)dx) = 0`

but as the two quantities are non-negative, this implies they are both null and then for every `xi in [a,b]`:

`int_a^xi f(x)dx = 0`

Differentiating with respect to `xi` and applying the fundamental theorem of calculus:

`d/(d xi)(int_a^xi f(x)dx) = f(xi) = 0`