I want to show that if the integral offwith upper limitband lower limitaisequal to 0

thenf(x) = 0for allx in [a,b]?

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- Nov 14th 2006, 12:41 PMdopiintegral 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]?** - Nov 14th 2006, 01:06 PMPlato
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}$ - Nov 14th 2006, 01:07 PMThePerfectHacker
Which is clearly not true,

$\displaystyle f(x)=x$ the identity function on $\displaystyle [-1,1]$ :eek:

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. - Nov 14th 2006, 01:13 PMSoroban
Hello, dopi!

Quote:

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

- Nov 14th 2006, 01:14 PMdopi

**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" - Nov 14th 2006, 01:23 PMThePerfectHacker
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!) - Nov 14th 2006, 01:26 PMdopi