1. ## integral question

I want to show that if the integral of f with upper limit b and lower limit a is equal to 0

then f(x) = 0 for all x in [a,b]?

2. That is not true. Here is a counter-example.
$\begin{array}{l}
x \in \left[ {0,1} \right]\quad \& \quad f\left( x \right) = \left\{ {\begin{array}{lr}
{\frac{1}{n}} & {x = \frac{1}{n}} \\
0 & \mbox{else} \\
\end{array}} \right. \\
\int\limits_0^1 {f(x)dx} = 0 \\
\end{array}$

3. Originally Posted by dopi
I want to show that if the integral of f with upper limit b and lower limit a is equal to 0

then f(x) = 0 for all x in [a,b]?
Which is clearly not true,
$f(x)=x$ the identity function on $[-1,1]$

Perhaps, you wish to show that if $f(x)\geq 0$
And if, $\int_a^b f(x)dx=0$
Then, $f(x)=0$

But that is also not true.
Consider an analoge of the Dirac Delta Function.
That is, the function is discontinous at one point with a different value and elsewhere is equal to zero.

4. Hello, dopi!

I want to show that if $\int^b_a f(x)\,dx \;=\;0$, then $f(x) = 0$ for all $x \in [a,b]$

You are saying:
. . If the area 'under a curve' on [a,b] is zero, the curve must be the x-axis.

This is not true . . .

Counter-examples: . . $\int^a_{\text{-}a}x\,dx\qquad\quad\int^{2\pi}_0\sin x\,dx\qquad\quad\int^b_{\text{-}b}(x^3 - 4x)\,dx$

5. Originally Posted by ThePerfectHacker
Which is clearly not true,
$f(x)=x$ the identity function on $[-1,1]$

Perhaps, you wish to show that if $f(x)\geq 0$
And if, $\int_a^b f(x)dx=0$
Then, $f(x)=0$

But that is also not true.
Consider an analoge of the Dirac Delta Function.
That is, the function is discontinous at one point with a different value and elsewhere is equal to zero.

yea thats wat i wanted to show if f(x)>0 and if the integral of f= 0 with upper limit b and lower limit a then f(x)= 0 for all x in [a,b]...but i dont get what this has to do with the qiestion "Consider an analoge of the Dirac Delta Function.
That is, the function is discontinous at one point with a different value and elsewhere is equal to zero"

6. Originally Posted by dopi
yea thats wat i wanted to show if f(x)>0 and if the integral of f= 0 with upper limit b and lower limit a then f(x)= 0 for all x in [a,b]...but i dont get what this has to do with the qiestion "Consider an analoge of the Dirac Delta Function.
That is, the function is discontinous at one point with a different value and elsewhere is equal to zero"
Consider the function,
$f(x)=0$ on $[-1,0)\cup (0,1]$ and $f(x)=1$ for $x=0$
Then, the integral still exists,
$\int_{-1}^1 f(x)dx=0$ though it is discontinous on this interval.

Yet,
$f(x)\not = 0$ for all the points in $x\in [-1,1]$

Perhaps, you want to show $f(x)\geq 0$ and $f(x)$ is continous on the interval. (Look how many conditions you have to fullfil to demonstrate what you have said!)

7. Originally Posted by ThePerfectHacker
Consider the function,
$f(x)=0$ on $[-1,0)\cup (0,1]$ and $f(x)=1$ for $x=0$
Then, the integral still exists,
$\int_{-1}^1 f(x)dx=0$ though it is discontinous on this interval.

Yet,
$f(x)\not = 0$ for all the points in $x\in [-1,1]$

Perhaps, you want to show $f(x)\geq 0$ and $f(x)$ is continous on the interval. (Look how many conditions you have to fullfil to demonstrate what you have said!)
thankz..sori bout that ..im jus going to confuse my self even more...thankz for the help