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Will the Fundamental Theorem of Calculus be useful in evaluating:

$$y=\int_{0.5}^{1} \left \lfloor t \right \rfloor dt?$$

The Riemann sums make this evaluation easy. I, however, would like to know how( if possible), can I use the theorem to evaluate this definite integral. Here, I assumed that $y$ is continuous in $[0.5,1]$ and differentiable in $(0,1)$.

Edit: To be clear about what I wish to know, I ask if I could evaluate the definite integral just by taking the difference, $y(1)-y(0.5)$( second part of the Fundamental Theorem of Calculus), where $y$ is the antiderivative of $\left \lfloor x\right \rfloor$.

I'm not sure if I'm allowed to talk about antiderivatives of discontinuous functions. Am I allowed to?

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  • $\begingroup$ It's left continuous at 1, but I don't think we can use FTC still. Seems a little sketch, but good question. $\endgroup$ – Sean Roberson Jun 24 '17 at 3:19
  • $\begingroup$ Don't you think the floor function is right continuous at 1? $\endgroup$ – R004 Jun 24 '17 at 3:30
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    $\begingroup$ Well, yes, but it's certainly not two-sided continuous. $\endgroup$ – Sean Roberson Jun 24 '17 at 3:31
  • $\begingroup$ Normally the textbooks of calculus present the Fundamental Theorem of calculus for continuous functions only. It's great that you want to know if these theorems hold for discontinuous functions or not +1. I have given a link in my answer which deals with the general Fundamental Theorem of calculus which is applicable for Riemann integrable functions. $\endgroup$ – Paramanand Singh Jun 24 '17 at 8:45
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I would suggest you to have a look at this answer where I have discussed the Fundamental Theorem of Calculus for functions which are not necessarily continuous.

Next note that the integrand here has a discontinuity at $x=1$ and as far as the interval of integration is concerned the discontinuity is removable and hence we just remove it by changing value of integrand at $x=1$ to $0$. Doing this has no impact on the value of the integral (because the value of a Riemann integral does not depend on the value of the integrand at a finite number of points in the interval under consideration) and thus the desired integral is equal to $\int_{1/2}^{1}0\,dx=0$.


In general if we have to evaluate an integral of the form $\int_{a} ^{b} f(x) \, dx$ where $f$ has a finite number of discontinuities at $x_{1},x_{2},\dots,x_{n}$ in $[a, b] $ with $$a\leq x_{1}\leq x_{2}\leq \dots\leq x_{n} \leq b$$ such that left and right hand limits of $f$ exist at each point of discontinuity then we split the integral as follows $$\int_{a} ^{b} f(x) \, dx=\int_{a} ^{x_{1}}f(x)\,dx+\int_{x_{1}}^{x_{2}}f(x)\,dx+\cdots+\int_{x_{n-1}}^{x_{n}}f(x)\,dx+\int_{x_{n}}^{b}f(x)\,dx$$ Each integral on right can be evaluated by observing that the integrand has removable discontinuity at the end points $x_{i}$ and the integrand can be redefined at these points to make it continuous on that interval.

You can use this technique to evaluate $\int_{1}^{2}[x^{2}]\,dx$ as $$\int_{1}^{2}[x^{2}]\,dx=\int_{1}^{\sqrt{2}}1\,dx+\int_{\sqrt{2}}^{\sqrt{3}}2\,dx+\int_{\sqrt{3}}^{2}3\,dx$$


Also to answer your question about evaluating the integral under consideration via second part of fundamental theorem of calculus, note that there is no anti-derivative of $[x] $ on interval $[1/2,1]$ (why? perhaps you should answer this yourself, but let me know if you feel issue here) and hence we can't use fundamental theorem of calculus here.

Also note that continuous functions are guaranteed to have an anti-derivative, but continuity is not necessary. There are discontinuous functions which possess anti-derivative and if we have a function $f$ which is Riemann integrable on $[a, b] $ and possesses an anti-derivative $F$ (note that existence of anti-derivative is not guaranteed and hence we are assuming its existence here as a part of the hypotheses) such that $F'(x) =f(x) $ for all $x\in[a, b] $ then we have $\int_{a}^{b}f(x)\,dx=F(b)-F(a)$.

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  • $\begingroup$ I might know why $[x]$ has no antiderivative in the intervals containing integers. I will use the method of contradiction to prove this. Let us say that the step function does have an antiderivative. If we sketch its antiderivative, we will find sharp turns at integers, suggesting that the derivatives at those points are not defined. But $[x]$ has unique values at integer points. This is absurd. $\endgroup$ – R004 Jun 24 '17 at 8:59
  • $\begingroup$ @R004: Your intuitive argument is correctbut this can not be used as a proof. You need to prove that derivatives can not have jump discontinuity via mean value theorem. Thus a step function like $[x]$ with jumps can not be the derivative of any function. $\endgroup$ – Paramanand Singh Jun 24 '17 at 9:23
  • $\begingroup$ @R004: See one proof here in this answer : math.stackexchange.com/a/577978/72031 $\endgroup$ – Paramanand Singh Jun 24 '17 at 9:35
  • $\begingroup$ If I understand this correctly, what $\int_{1/2}^1 [x]dx=\lim_{t \to 1^-} \int_{1/2}^t [x]dx$ is telling me is that I can carry out my regular integration using the SFTC, right? $\endgroup$ – R004 Jun 24 '17 at 10:18
  • $\begingroup$ @R004 : this statement is saying that the function $y(x) =\int_{1/2}^{x}[t]\,dt$ is continuous at $x=1$. It does not have any other meaning. By the way if you check the statement of first FTC in my linked answer you will find that it says : If $F(x) =\int_{a} ^{x} f(t) \, dt$ then $F(x) $ is continuous on $[a, b] $. So this is just a consequence of first FTC. $\endgroup$ – Paramanand Singh Jun 24 '17 at 10:21
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$y $ is continuous at $[0.5,1] \implies $

$$\int_{0,5}^1\lfloor t\rfloor dt= \lim_{x\to 1^-}\int_{0,5}^x\lfloor t \rfloor dt =0. $$

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  • $\begingroup$ I have started learning recently. I will need you to elaborate on this if you don't mind. $\endgroup$ – R004 Jun 24 '17 at 3:26
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By definitión $[x]$ is defined as the biggest integer notexceding $x$, for example $[\pi]=3$ and $[-1/2]=-1$. The floor function $x \mapsto [x]$ is continuous in the interval $[1/2,1)$; so $$\int_{1/2}^1 [x]dx=\lim_{t \to 1^-} \int_{1/2}^t [x]dx.$$ Also, for $1/2\leq x<1$ we have that $[x]=0$, so $$\int_{1/2}^1 [x]dx=\lim_{t \to 1^-} \int_{1/2}^t [x]dx=\lim_{t \to 1^-} \int_{1/2}^t 0dx=\lim_{t \to 1^-}0=0$$

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    $\begingroup$ Ohhh you mean Barrow's Rule, sorry. For your firs question, it is just the definition of limit. For example, consider the function $F(x)=\int_1^x t^2dt$, you know by the second part of the Fundamental Theorem of Calculus (Barrow's Rule) that $F(x)=\frac{x^3}{3}-\frac{1}{3}$, so for example $$\lim_{x \to 0} F(x)=\lim_{x \to 0} \int_1^x t^2dt=\lim_{x \to 0}(x^3/3-1/3)=-1/3.$$ As for your second question, yes; it is possible. Just have in mind that the antiderivat of $[x]$ is $g(x):=x[x]-\frac{1}{2}[x]([x]+1)$. Then you may use Barrow's Rule without problem. Hope this helps! $\endgroup$ – Adrián Naranjo Jun 24 '17 at 3:48
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    $\begingroup$ Because by the limits of integration it follows that $1/2\leq x \leq 1$, note that $[x]=0$ if $1/2 \leq x <1$ and $[x]=1$ when $x=1$, so we will have "two posible values for [x]". To avoid this problem I use the fact that [x] is continous on the interval we are working on, so I can take limits without any problem. $\endgroup$ – Adrián Naranjo Jun 24 '17 at 3:58
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    $\begingroup$ Yes, let's say it is more "natural". $\endgroup$ – Adrián Naranjo Jun 24 '17 at 4:05
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    $\begingroup$ I think you should also mention that if $f$ is Riemann integrable on $[a, b] $ and $F(x) =\int_{a} ^{x} f(t) \, dt$ then $F(x) $ is continuous on $[a, b] $. Because of this result $y(x) =\int_{1/2}^{x}[t]\,dt$ is continuous and hence $$\int_{1/2}^{1}[x]\,dx=y(1)=\lim_{x\to 1^{-}}y(x)=\lim_{x\to 1^{-}}\int_{1/2}^{x}[t]\,dt$$ Without this your first equation seems to come from out of the blue. $\endgroup$ – Paramanand Singh Jun 24 '17 at 7:14
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    $\begingroup$ Also you mention in comments about anti-derivative of $[x] $ and provide a formula for it. This is wrong because the function $[x] $ does not have an anti-derivative over any interval which includes integer points. I wonder how you arrived at a formula for anti-derivative when it does not exist. $\endgroup$ – Paramanand Singh Jun 24 '17 at 8:51
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Yes you can talk about the anti derivatives of discontinuous functions. But if a function $f$ has a jump discontinuity (like the greatest integer function $[x] $ in question which has jump discontinuity at integers) then it does not possess an anti-derivative and hence it's integral can not evaluated by using Leibniz Rule (that is difference between values of anti-derivative at the end points of the interval).

There are discontinuous functions which don't have jump discontinuity and then they may possess anti-derivative. For example the function $f(x)=2x\sin(1/x)-\cos(1/x),f(0)=0$ is continuous everywhere except at $0$. It possesses an anti-derivative $g(x) =x^{2}\sin(1/x),g(0)=0$ and for all real $a, b$ we have $$\int_{a} ^{b} f(x) \, dx=g(b) - g(a) $$ However in cases where the integrand is having jump discontinuity in the interval of integration there is still some hope. Here we can not use anti-derivative (because it does not exist) but we can split the interval of integration into multiple intervals via the points of discontinuity and evaluate integral on each subinterval and add these integrals to get the value of the integral on the given interval. Splitting the interval of integration into multiple sub-intervals has the advantage that the function has discontinuity only at the end points and these are removable. So the function can be modified accordingly at end points to make it continuous and the integral evaluated as usual.

For the current question the integrand already has discontinuity on end point of the interval so no need to split the interval and integral is directly evaluated as $\int_{1/2}^{1}0\,dx=0$.

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  • $\begingroup$ In order to possess an anti-derivarive it is not necessary that the function be continuous. For example $f(x) =2x\sin(1/x)-\cos(1/x),f(0)=0$ has a discontinuity at $x=0$ but it has an anti-derivarive $g(x) =x^{2}\sin(1/x),g(0)=0$. However it is necessary that a function must not have jump discontinuity in order to possess an anti-derivarive. $\endgroup$ – Paramanand Singh Jun 24 '17 at 7:04
  • $\begingroup$ Yes but I was talking about evaluating definite integrals. $\endgroup$ – vishal mishra Jun 24 '17 at 7:10
  • $\begingroup$ OK I get it now. I didn't mention this explicitly in my answer. $\endgroup$ – vishal mishra Jun 24 '17 at 7:11
  • $\begingroup$ Again you are wrong. The evaluation of integral as a difference of values of anti-derivarive at the end points of the interval has nothing to do with the integrand being continuous in the interval. All that we need is that the integrand be Riemann integrable in the interval (and not necessarily continuous). $\endgroup$ – Paramanand Singh Jun 24 '17 at 7:21
  • $\begingroup$ I don't think so I will give you a example.integerate 1+x^2 from 0 to 3^1/2. Write the anti derivative as 1/2arctan(2x/1-x^2) $\endgroup$ – vishal mishra Jun 24 '17 at 7:25

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