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]?
That is not true. Here is a counter-example.
$\displaystyle \begin{array}{l}
x \in \left[ {0,1} \right]\quad \& \quad f\left( x \right) = \left\{ {\begin{array}{lr}
{\frac{1}{n}} & {x = \frac{1}{n}} \\
0 & \mbox{else} \\
\end{array}} \right. \\
\int\limits_0^1 {f(x)dx} = 0 \\
\end{array}$
Which is clearly not true,
$\displaystyle f(x)=x$ the identity function on $\displaystyle [-1,1]$
Perhaps, you wish to show that if $\displaystyle f(x)\geq 0$
And if, $\displaystyle \int_a^b f(x)dx=0$
Then, $\displaystyle f(x)=0$
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.
Hello, dopi!
I want to show that if $\displaystyle \int^b_a f(x)\,dx \;=\;0$, then $\displaystyle f(x) = 0$ for all $\displaystyle x \in [a,b]$
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: . . $\displaystyle \int^a_{\text{-}a}x\,dx\qquad\quad\int^{2\pi}_0\sin x\,dx\qquad\quad\int^b_{\text{-}b}(x^3 - 4x)\,dx$
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,
$\displaystyle f(x)=0$ on $\displaystyle [-1,0)\cup (0,1]$ and $\displaystyle f(x)=1$ for $\displaystyle x=0$
Then, the integral still exists,
$\displaystyle \int_{-1}^1 f(x)dx=0$ though it is discontinous on this interval.
Yet,
$\displaystyle f(x)\not = 0$ for all the points in $\displaystyle x\in [-1,1]$
Perhaps, you want to show $\displaystyle f(x)\geq 0$ and $\displaystyle f(x)$ is continous on the interval. (Look how many conditions you have to fullfil to demonstrate what you have said!)