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it seems obvious that this integral is zero and so is the limit but what theorem we are using here?

I see it's connected to Riemann sums with an interval=zero Right ?

The function $\mathrm{f}$ is continuous.

$$\lim_{x \to 0}\int_0^x\mathrm{f}(x)\ \mathrm{d}x= \ ?$$

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Perhaps the Dirac delta function is an $f$ which gives a non-zero value – James Feb 4 '14 at 18:41
but if $\int_0^0 \delta=1$ then we also have $\int_0^0 \delta=\int_0^0 \delta+\int_0^0 \delta = 2$ – Sean D Feb 4 '14 at 23:47
the Dirac Delta is not continues so integration won't give me zero . that's if an integration existed for this function. – Was Fr Feb 5 '14 at 7:41
up vote 7 down vote accepted

We are not using any theorem. The definition of the definite integral is $$ \int_a^b f(x) \; dx = \lim_{n\to \infty} \sum_{i=1}^n f(x_i)\Delta x. $$ where $x_i = a + i\Delta x$ and $\Delta x = \frac{b - a}{n}$. If $a=b=0$, then $\Delta x = 0$ and so the integral is zero: $$ \int_0^0 f(x)\; dx = \lim_{n\to \infty}\sum_{i=1}^n 0 = \lim_{n\to \infty} 0 = 0. $$

About the limit. Assume that $f$ is continuous on a small interval $[0, \epsilon]$. Then according to the Fundamental Theorem of Calculus the function given by $$ F(x) = \int_0^x f(t) \; dt $$ is continuous on $[0,\epsilon]$. In particular $F$ is continuous at $0$. This, by definition, means that $\lim_{x\to 0^+} F(x) = 0$, or that $$ \lim_{x\to 0^+} \int_0^x f(t) \; dt = 0. $$

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yes seems a good explanation thank you. – Was Fr Feb 4 '14 at 17:56
but how can you explain that the integral is from 0 to 0 (can we use x=0 directly no problem because of the limit ?) – Was Fr Feb 4 '14 at 17:59
The key point in Thomas's argument is that the function $F(x)= \int_0^x f(x)dx$ is a continuous function (assuming f is Riemann integrable, f doesn't even need to be continuous, for F to be continuous). If some function g is continuous at $a$, $\lim_{x -> a} g(x) = g(a)$ by the definition of continuity. The function g here is the whole integral, i.e. F(x). – MHH Feb 4 '14 at 22:56

Assuming $f$ is Riemann integrable, it is bounded by some $B$, so $0 \le |\int_0^x f(x) dx | \le \int_0^x |f(x)| dx \le Bx$.

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i am interested when apply equal sign?or less sign? – dato datuashvili Feb 4 '14 at 17:52
@datodatuashvili: I don't understand your question. In the above, if you let $f(x) = B = 0$ for all $x$, then the inequalities are equalities. – copper.hat Feb 4 '14 at 17:54
How did you deduce that B=0 ? – Was Fr Feb 4 '14 at 18:04
It is just an example to show that the $\le$ cannot be replaced by $<$ as @datodatuashvili had asked. In general, $B$ is not zero. $B$ is just a bound on the function $f$. – copper.hat Feb 4 '14 at 18:06
oh yes I agree. and adding more info : max(f(x))*(b-a) is also a bound. an upper bound – Was Fr Feb 4 '14 at 18:21

Another approach valid when $f$ is continuos: the integral is a function $F$ s.t. $F′=f$ and $F(0)=0$. By continuity, $\lim F=F(0)=0$.

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but who said that F(0)=0 ? – Was Fr Feb 4 '14 at 17:55
Integral from 0 to... 0. – Martín-Blas Pérez Pinilla Feb 4 '14 at 17:57
I agree that it is from 0 to 0 so : integral(fx)...= F(0)-F(0) but we don't know the value of F(0) – Was Fr Feb 4 '14 at 18:02
F(0) could be 100 and still the integral is zero . I think now I understand your way : F(0)-F(0)= A-A=0 done. I'm just reluctant about the fact that you said F(0)=0 Could you explain more about the curve ? Have a good day – Was Fr Feb 4 '14 at 18:06
My $F(x)=\int_0^x f(t)dt$ by definition (Fundamental Theorem of Calculus). You are thinking in the Barrow's Rule. – Martín-Blas Pérez Pinilla Feb 4 '14 at 18:10

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