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I was studying integrals and just out of curiosity,

Does there exist any 'continuous' functions such that $\int_a^af(x) \, dx$ ($a$ is any number) equals a value other than $0$?

Since continuous functions are Riemann integrable, so I think it should be $0$. Is this correct?

Also, with out the condition 'continuous', does there exist any function such that $\int_a^af(x) \, dx$ isn't $0$?

EDIT

I'm looking for any function that $\int_a ^a\ f(x) \neq 0$. Can anyone find me one?

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  • $\begingroup$ (First) Fundamental theorem of Calculus... $\endgroup$ – imranfat Jul 9 '16 at 0:26
  • $\begingroup$ @imranfat Okay, so the value of the integral should be 0 when the function is continuous. What about discontinuous ones? $\endgroup$ – zxcvber Jul 9 '16 at 0:31
  • $\begingroup$ If you use lebesgue integral and the measure is the counting measure, the answer is yes. $\endgroup$ – Xianjin Yang Jul 9 '16 at 0:31
  • $\begingroup$ @zxcvber It is always zero when the function $f$ is integrable. A more interesting question is something like $\int_0^0 \frac 1x \ dx$, which has been asked before (though I can't seem to find it now). $\endgroup$ – AlohaSine Jul 9 '16 at 0:32
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    $\begingroup$ @zxcvber To your question: If you know any measure theory, then define $\mu$ to be counting measure, which means for any set $A \subseteq \Bbb R$, $\mu(A)$ (the measure of $A$) is the number of elements in $A$. So the interval $[a,a]$ has one element, so $\mu([a,a]) = 1$. Now, take $f(x) =1$ for every $x \in \Bbb R$. So $f$ is the constant function $1$. Then $\int \limits_{a}^{a} f(x) \,d\mu = 1$. Notice the $dx$ from earlier is replaced with $d\mu$ here because $dx$ usually means the integral is with respect to Lebesgue measure. $d\mu$ means it is with respect to the measure $\mu$. $\endgroup$ – layman Jul 9 '16 at 1:11
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No, for $f(x)$ any continuous function, $\int_a^a f(x)dx= 0$. As for non-continuous functions, my first thought was a "Dirac delta function" for which $\int_C \delta(x)dx= 1$ for any set, C, containing 0. However, a "delta function" is not a true function- it is a "generalized function" or "distribution".

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  • $\begingroup$ So, $\int_0^0 \delta (x) dx$ equals $1$?, and why is a "generalized function" not a true function? $\endgroup$ – zxcvber Jul 9 '16 at 0:36
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    $\begingroup$ @zxcvber There is no function with the properties that $\delta$ has, so it's not a function. The notation $\int \delta(x)f(x)\mathrm{d}x$ is misleading to the uninitiated, since it's not defined as an integral at all, but the application of a linear functional called delta to the function $f(x)$. $\endgroup$ – arctic tern Jul 9 '16 at 0:40
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    $\begingroup$ The delta function can sort of be a counterexample if you think of it as a measure instead of a distribution. Consider the Lebesgue integral $\int_0^01d\delta:=\int_{\{0\}}1d\delta=1$. $\endgroup$ – Funktorality Jul 9 '16 at 1:01
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You are right, when talking about the Riemann Integral, then $$ \int_a^a f(x)\; dx $$ always equal to zero. If this integral is defined, then it is zero. To see this, you just have to write out the definition of the integral: $$ \int_a^a f(x)\; dx = \lim_{n\to \infty}\sum_{i=0}^{n} f(x_i^*) \, \Delta x $$ Here $\Delta x = \frac{a - a}{n} = 0$, so all the sums will be zero, and so the limit will be zero.

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