What is the function that satisfies $\int_0^x f(t) dt=constant$ [closed]

$$\int_0^x f(t) dt=constant$$

What is the function that satisfies this condition ?

Thank you!

closed as off-topic by Travis Willse, Aaron Maroja, A.P., Harish Chandra Rajpoot, graydadNov 3 '15 at 2:49

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• Is $\int_0^x f(t) dt$ a function of $x$? – Mythomorphic Nov 2 '15 at 17:30
• yes! a function of x, f(x) – user286255 Nov 2 '15 at 17:35

Differentiating both sides of $$\int_0^x f(t) dt=constant$$ using Fundamental Theorem Of Calculus, as mentioned by Omnomnomnom in his comment,( or Leibnitz Rule as I had earlier written), we get $$f(x)=0$$

• This is more commonly referred to as the "fundamental theorem of calculus" than the "Leibniz Rule" – Omnomnomnom Nov 2 '15 at 17:39
• @Omnomnomnom The above result can be arrived at using both the "fundamental theorem of calculus" and the "Leibniz Rule". – SchrodingersCat Nov 2 '15 at 17:41
• sure, and the product rule can be derived using the multivariable chain-rule. However, since the asker is likely an introductory calculus student, it is better to use the more accessible result. – Omnomnomnom Nov 2 '15 at 17:57
• I agree. You're right. – SchrodingersCat Nov 2 '15 at 17:59
• OP hasn't ever specified that $f$ is continuous. If it isn't, you can't apply the fundamental theorem of calculus. – Wojowu Nov 2 '15 at 18:29

Let's denote the continuous function:

$$G(x)=\int_0^x f(t) dt$$

Then by the Fundamental Theorem of Calculus,

$$\frac{d}{dx}G(x)=f(x)$$

When $G(x)=C$, $$\frac{d}{dx}G(x)=0=f(x)$$

• OP hasn't ever specified that $f$ is continuous. If it isn't, you can't apply the fundamental theorem of calculus. – Wojowu Nov 2 '15 at 18:29

Because OP has never said $f(t)$ has to be a continuous function, there are more possible functions than a constant zero. For example, if we define $$f(t)=1\text{ for t\in\Bbb Z}, 0\text{ otherwise}$$ We get a nonconstant function integral of which is $0$ on every integral. More generally, any function which is non-zero on a discrete set of points will work.

But these aren't all functions: take, for example, Thomae's function. It is only nonzero on a set of measure zero, and is Riemann-integrable, hence its integral over any interval is again zero.

If we limit ourselves to Riemann-integrable functions, I believe this is the necessary and sufficient condition:

$f(t)=0$ for every $t$ at which $f$ is continuous.

Since Riemann-integrable function must be continuous outside a set of measure zero, this condition is enough to conclude $f$ has integral $0$ on every interval. Conversely, if $f(t_0)>0$ at a point of continuity $t_0$ ($f(t_0)<0$ is treated similarly), then on some open interval we have $f(t)>\frac{f(t_0)}{2}>0$, so on this interval the integral will increase, so it won't be constant.

If we allow more general integrals (Lebesgue measurable) then I believe an equivalent condition would be that $f(t)$ is non-zero only on a set of measure zero, but I don't know enough about Lebesgue-integrability to say that with certainty.

Clearly the constant is zero,Because:

$$constant = \int_0^0 f(x) dx = 0$$ Can you conclude now?

• @SimonS: on the opposite ! It shows that such a function does not exist, unless the constant is $0$. – Yves Daoust Nov 2 '15 at 17:41
• @YvesDaoust Yes, my mistake – Simon S Nov 2 '15 at 18:06
• if constant is not zero and x>0, what about this case? – user286255 Nov 2 '15 at 18:06
• @user286255: this is answered here. – Yves Daoust Nov 2 '15 at 18:18