Integration

Question

$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)

Comments

Graph of which function ?

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now check

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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) = 1/0^2/7 = 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..

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@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

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@srestha ,

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$f(x)=\int_{-2}^{2}x^{-2/7}dx=k$, 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.

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@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

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@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 ?

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wait I will give u all graphs

but still not getting , only with those 3 points how u get this graph

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@ankit

for this question graph will be like this.

Where is it discontinuous?

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@srestha , can you please tell me the value of this function  at x = 0 in this graph ?

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it will be 0 for both the limit (I mean for LHL and RHL both)

right?

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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

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@Ankit

I too thought it should be discontinuous

but look at graph here

https://www.desmos.com/calculator/lijenxf85n

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@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..

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@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. 

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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. 

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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...    

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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. 

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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.

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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

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@ 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.

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And another point, what integration means?

It means the discrete region(not continuous region) where u checking existence of graph

right? 

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@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.

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@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 ? 

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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. 

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is not corrct.

Why not correct? 

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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. 

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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

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$\int_{a}^{b} f(x)$ is discrete function while$\sum_{a}^{b}f(x)$continuous

See the picture given here

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@ 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

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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 ...  

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∫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. 

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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.

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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.  

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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.  

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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 $b$  "

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@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]

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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.. 

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@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.  

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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?"