0 votes 0 votes $f\left ( x \right )=$$\int_{-2}^{2}x^{-\frac{2}{7}}dx$ Is this function f(x) is continuous, bounded and differentiable? (In exam hall is it possible to draw the graph for this function f(x), or some other procedure to follow to ans this) Calculus integration + – srestha asked May 18, 2018 • edited May 18, 2018 by srestha srestha 1.8k views answer comment Share Follow See all 35 Comments See all 35 35 Comments reply Kushagra Chatterjee commented May 18, 2018 reply Follow Share Graph of which function ? 0 votes 0 votes srestha commented May 18, 2018 reply Follow Share now check 0 votes 0 votes ankitgupta.1729 commented May 18, 2018 reply Follow Share here ,f(x) = x-2/7 now ,f(-x) = (-x)-2/7 = ((-x)2)-1/7 = (x2)-1/7 = x-2/7 So, here f(-x) = f(x)...So, f(x) is an even function..now we can use property of definite integral here.. for drawing the graph :- since f(x) is an even function..so it will be symmetric about Y-axis...or we can say, Y- axis work as a mirror..now make the graph for negative values of x..and then reflect it about Y-axis.. at x=0 , f(x) = 1/02/7 = tends to infinity.. so f(x) will be tends to infinity at x =0.. and I don't think any other trick will work to make this graph without taking negative x points for this graph.. 0 votes 0 votes srestha commented May 18, 2018 reply Follow Share @ankit then how the graph look like? what I got a function is bounded, if it is continuous So, if we can prove it is continuous, we can say it is bounded and continuous, which serves our 2 purposes :) https://math.stackexchange.com/questions/1608078/sufficient-conditions-for-bounded-function 0 votes 0 votes ankitgupta.1729 commented May 18, 2018 reply Follow Share @srestha , 0 votes 0 votes joshi_nitish commented May 18, 2018 reply Follow Share $f(x)=\int_{-2}^{2}x^{-2/7}dx=k$, where k is some constant(since it is definite integral with constant definite limits). now since f(x) is constant throughout it's domain therefore it is continuous and differentiable and also bounded. 1 votes 1 votes srestha commented May 18, 2018 reply Follow Share @nitish only that point can say graph is continuous? compare it with https://gateoverflow.in/8124/gate2015-2_26 @ankit isnot the graph overlapping in x axis? I have got this graph as continuous 0 votes 0 votes ankitgupta.1729 commented May 18, 2018 reply Follow Share @srestha , could you please explain why it is continuous in domain [-2,2]... according to you, graph from left and right side will meet at infinity at point x=0 ? 0 votes 0 votes srestha commented May 18, 2018 reply Follow Share wait I will give u all graphs but still not getting , only with those 3 points how u get this graph 0 votes 0 votes srestha commented May 19, 2018 reply Follow Share @ankit for this question graph will be like this. Where is it discontinuous? 0 votes 0 votes ankitgupta.1729 commented May 19, 2018 reply Follow Share @srestha , can you please tell me the value of this function at x = 0 in this graph ? 0 votes 0 votes srestha commented May 19, 2018 reply Follow Share it will be 0 for both the limit (I mean for LHL and RHL both) right? 0 votes 0 votes srestha commented May 19, 2018 reply Follow Share good resource of continuity and differentiability https://www.khanacademy.org/math/calculus-home/limits-and-continuity-calc/continuity-review-calc/e/continuity-challenge https://www.youtube.com/watch?v=dhRMPlOB7ec 1 votes 1 votes srestha commented May 19, 2018 reply Follow Share @Ankit I too thought it should be discontinuous but look at graph here https://www.desmos.com/calculator/lijenxf85n 0 votes 0 votes ankitgupta.1729 commented May 19, 2018 reply Follow Share @srestha ,here LHL=RHL=f(0) = infinity.. when x tends to zero , then values of f(x) tend to infinity..so here from both sides of X-axis , we are going to +infinity...and as values of x will go towards zero then values of f(x) will be increasing each time..so it will be increasing from both sides of X-axis... So, intuitively , we can say that they will meet at infinity somewhere..so it will be continuous... but infinity is not a finite number and it is not defined and intuition could be wrong also..so I don't know they will meet or not.. 0 votes 0 votes Kushagra Chatterjee commented May 19, 2018 reply Follow Share @srestha read this http://math.mit.edu/~jspeck/18.01_Fall%202014/Supplementary%20notes/01c.pdf there you will know about 4 types of discontinuity. As mentioned by Ankit at x = 0 the function given goes to infinity so it is of the type infinite discontinuity. 1 votes 1 votes Deepak Poonia commented May 19, 2018 reply Follow Share since it is definite integral with constant definite limits now since f(x) is constant throughout it's domain therefore it is continuous and differentiable and also bounded. Agreeing with Nitish. Definite integral of any function with constant definite limits will give Area under that curve, which will be some constant value and hence a straight line parallel to $x-axis$ . So, All the Graphs etc being drawn in other comments are not for the $f(x)$ you originally defined in the question. 0 votes 0 votes ankitgupta.1729 commented May 19, 2018 reply Follow Share since it is definite integral with constant definite limits. now since f(x) is constant throughout it's domain therefore it is continuous and differentiable and also bounded. Not agreeing with Nitish. counter-example :- if f(x) = |x| and limits are from -1 to +1 then area under curve is a constant value but f(x) is not differentiable at x=0... 0 votes 0 votes Deepak Poonia commented May 19, 2018 reply Follow Share f(x) = |x| and limits are from -1 to +1 then area under curve is a constant value but f(x) is not differentiable at x=0... $g(x) = \int_{-2}^{2}\left | x \right |dx$, So, What you are arguing is for $f(x) = \left | x \right |$, Not for $g(x)$. $g(x)$ is continuous,differentiable etc. In the Original Question, Just focus on the RHS and solve it once, You will get some value, say, $k$. So, after that the Original equation will become $f(x) = k $. now, the graph of $f(x)$ is a Line. 0 votes 0 votes ankitgupta.1729 commented May 19, 2018 reply Follow Share I took f(x) = x-2/7 because I assumed that she was asking to draw the graph of f(x) because most of the people know that definite integral gives area under curve which is a constant value ..so it will become easy to say about continuity and differentiability about f(x) = constant....so I thought nitish was talking about x-2/7 0 votes 0 votes srestha commented May 20, 2018 reply Follow Share @ Deepakk graph is drawn only for function, not for integration. Still it is showing continuous. So, What you are arguing is for f(x)=|x|, Not for g(x), g(x) is continuous,differentiable etc. is not corrct. ------------------------------------------------------------------------------ And another point, what integration means? It means the discrete region(not continuous region) where u checking existence of graph right? 0 votes 0 votes joshi_nitish commented May 20, 2018 reply Follow Share @ankitgupta.1729 so I thought nitish was talking about x-2/7 is $x^{-2/7}$ a constant function? don't take self assumptions, it is harmful for GATE. 0 votes 0 votes ankitgupta.1729 commented May 20, 2018 reply Follow Share @joshi_nitish f(x)=k, where k is some constant do u think that this statement is true here since f(x) is unbounded and discontinuous at x=0 and tends to +infinity as x tends 0 ? 0 votes 0 votes Deepak Poonia commented May 20, 2018 reply Follow Share graph is drawn only for function, not for integration. I agree. But (talking about the$f(x)$ in the Question), $f(x)$ is not $x^{-2/7}$. But $f(x) = \int_{-2}^{2}x^{-2/7}dx$ And when we solve RHS, it becomes $f(x) = K$, So, in the Question, He is asking about properties of $f(x)$, Not of $x^{-2/7}$. So, If this is a Question of some exam then the answer must be given for $f(x) = K$, that's all that I am saying. is not corrct. Why not correct? And another point, what integration means? It means the discrete region(not continuous region) where u checking existence of graph No, Integration is always for continuous function. Summation is for Discrete functions. $\int vs \sum$. What $\sum$ does for Discrete functions, $\int$ does for Continuous functions. Integration (Mathematics), the computation of a definite integral, a fundamental concept of calculus, which allows, among many other uses, computing areas and averaging continuous functions. an integral (definite integral from a to b ) assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. 0 votes 0 votes Deepak Poonia commented May 20, 2018 reply Follow Share do u think that this statement is true here since f(x) is unbounded and discontinuous at x=0 What is $f(x)$ here. Kindly tell me. From what I can see in the Question, It is Some constant, Not $x^{-2/7}$. When Wherever they are interested in Graph of some functions in some interval, They write it so. They will ask you "$f(x)$ is continuous in $[2,4]$ or not" ; They won't give you $\int_{2}^{4}f(x)dx$. Like this Question @srestha shared in the comments : https://gateoverflow.in/8124/gate2015-2_26 0 votes 0 votes srestha commented May 20, 2018 reply Follow Share $\int_{a}^{b} f(x)$ is discrete function while$\sum_{a}^{b}f(x)$continuous See the picture given here 0 votes 0 votes srestha commented May 20, 2018 reply Follow Share @ Deepakk @nitish one thing tell me if $x^{-\frac{1}{3}}$ is discontinuous, then how $x^{-\frac{2}{7}}$ could be continuous? $\lim_{x\rightarrow 1-}\frac{1}{\left ( -1 \right )^{-2/7}}\neq \lim_{x\rightarrow 1+}\frac{1}{\left ( 1 \right )^{-2/7}}$ So, LHL not equal to RHL But only we draw graph showing it is not discontinuous 0 votes 0 votes ankitgupta.1729 commented May 20, 2018 reply Follow Share What is f(x) here. Kindly tell me. I was saying that we can't write f(x) = k , where 'k' is constant because function let's say g(x)=x-2/7 is discontinuous at x=0..so we can't find the area under curve within [-2,2] here because function x-2/7 is not bounded in [-2,2]..so area under curve will not give a finite constant value which can be defined ... 0 votes 0 votes Deepak Poonia commented May 20, 2018 i edited by Deepak Poonia May 20, 2018 reply Follow Share ∫baf(x)∫abf(x) is discrete function while∑baf(x)∑abf(x)continuous See the picture given here It's interesting to see the "Best answer" chosen answer. Interesting because it's wrong (Even the 2nd comment where He tries to derive it or Prove it, Is wrong). Comment number 5,9 are correct. Read the 9th one. Talking about that Question, In Calculus, For Monotonically Increasing Functions, We have certain results or relation between $\sum$ and $\int$ in a range, Which can be derived easily. It is given in any standard Calculus book like the one I referred long back "Apostol". You can find this result in Cormen too. In 1st or 2nd edition, you can find it in Chapter 3 I guess. In 3rd edition it is at the end in the Appendix. The results are as in the picture below : Refer Cormen 3rd edition "Appendix A Summations" for Proof of the above results. But the thing is that Do not derive from these Equalities that $\sum$ is continuous (Continuous addition) and $\int$ is discrete (Discrete addition). it's just the opposite. If you go through the Proofs of the above inequalities in Cormen, You will get what I'm trying to say. Hint for understanding the Proof : We know that $\sum_{m}^{n} f(x) = f(m) + f(m+1) + ..... + f(n)$ (look this is Discrete :) ) Think of $f(m)$ as $1 * f(m)$, then think of it as a Rectangle in which One side is of length 1 and other side is of length $f(m)$ , So, now you can think of $f(m)$ as the area of this rectangle. 0 votes 0 votes Deepak Poonia commented May 20, 2018 reply Follow Share if $x^{-1/3}$ is discontinuous, then how $x^{-2/7}$ could be continuous? Neither me nor Nitish is saying that $x^{-2/7}$ is continuous in the range [-2,2]. We are just saying $f(x)$ in the Question is Continuous. 0 votes 0 votes Deepak Poonia commented May 20, 2018 reply Follow Share I was saying that we can't write f(x) = k , where 'k' is constant because function let's say g(x)=x-2/7 is discontinuous at x=0..so we can't find the area under curve within [-2,2] here because function x-2/7 is not bounded in [-2,2]..so area under curve will not give a finite constant value which can be defined ... One of the Very Basic rules of the Integral Calculus " If limits of the definite integral are from $a$ to $b$ and the function (on which integration is being applied) is discontinuous at some point $c$ then the integration can be Broken down to "$a$ to $c$ + $c$ to $b$ " 0 votes 0 votes Kushagra Chatterjee commented May 20, 2018 reply Follow Share @Deepakk can u please show me how to compute the definite integral of x^(-2) from -2 to 2 by breaking it into [-2,0) and (0,2] 0 votes 0 votes ankitgupta.1729 commented May 20, 2018 reply Follow Share One of the Very Basic rules of the Integral Calculus " If limits of the definite integral are from a to b and the function (on which integration is being applied) is discontinuous at some point c then the integration can be Broken down to "a to c + c to b " here , it is the case of infinite discontinuity.. I don't think what are saying is applicable here.. 0 votes 0 votes Deepak Poonia commented May 20, 2018 reply Follow Share @Deepakk can u please show me how to compute the definite integral of x^(-2) from -2 to 2 by breaking it into [-2,0) and (0,2] Nice point. Thanks for pointing it out. Will certainly be worth digging about it. 0 votes 0 votes Deepak Poonia commented May 20, 2018 reply Follow Share here , it is the case of infinite discontinuity.. I don't think what are saying is applicable here.. Thank you @ankitgupta for mentioning. Will dig into it. And One long-pending doubt "How to tag someone on GO?" 0 votes 0 votes Please log in or register to add a comment.