Impossible integral question
By Murray Bourne, 10 Sep 2010
Scarlett, a reader of the IntMath Newsletter, recently wrote:
Can you help me with this question pls? Thx 🙂
Integration from 0 to -1
dxthis is what i manage to work out
let u = 2x^2+x
(du)^-2 = (4x+1)^-2 dx
1/2 int (u^2)/2 (du)^-2then i'm stuck bcoz i dunno how to solve (du)^-2
Her question says:
Her last line actually means this:
Since Scarlett was willing to put in some effort to help herself, I replied:
Hi Scarlett
Good on you for showing your working! (Most people don't and I have no idea where they are having a problem).
Interesting try, but you need to make sure you end up with "du" (never (du)^2) and the "du" needs to be on top of the fraction (you should never have (du)^-2).
Now, before going into the solution, let's point out something important.
Care with order of limits!
Scarlett wrote "Integration from 0 to -1".
This literally means the following (with the 0 as the lower limit of the integral and −1 as the upper limit).
But if we reverse the order of the limits, our final answer will be the negative of the answer without reversed limits.
Since most school math text books have the smallest (left-most) number as the lower limit, I presumed her question actually meant the following (with −1 as the lower, left-most limit and 0 as the upper limit).
This is the version we are going to solve.
This is a language issue, where we normally read top to bottom (Integration from 0 to -1) but we need to say integral limits from the bottom number to the top. It should have been Integration from -1 to 0.
Now, before I go on, have any of you spotted why I titled this article "Impossible integration question"?
Possible approaches
When you come across a new integral, you often need to do a bit of trial and error until you get the right approach.
Like Scarlett, I initially thought it might be a substitution, but with the more likely choice for u being:
u = 4x + 1
This will give u2 on the bottom and du on top, but there is a function of x on top (not a "nice" number), so it looks like it might give trouble.
Perhaps there is an easier way.
Partial Fraction approach
My next thought was to try partial fractions. This process allows us to split apart a fraction into its parts, thereby giving us a much simpler expression to integrate.
For simplicity, I first take the 1/2 out the front, remove it (and I need to remember to put it back at the end).
Since we have a denominator that contains repeated linear factors, we are looking for an expression in the form:
However, you can try all day and never solve this for A and B, since we can only find partial fractions if the degree of the numerator (highest power of x on top) is less than the degree of the denominator (highest power of x on the bottom).
In this example, the degree of both top and bottom is 2, so we cannot use this approach to finding the partial fractions.
However, with a bit of magic, we can still find a partial fraction decomposition and solve our integral.
[Note: Math teachers often seem to perform "magic" by pulling things out of thin air in order to solve some problem. I'm able to perform the following trick because I worked backward from the answer :-). It's worth following through the steps, though, because we often have to perform such tricks.]
Starting with the fraction:
We recall that (4x + 1)2 = 16x2 + 8x +1 and observe that the 2 terms on the top of the fraction bear some similarity to this expansion.
So we multiply top and bottom by 8:
Next, we add 1 to the top (to give the desired expression) and subtract one to preserve equality. On the bottom, I multiplied the 2 and 8 to give 16 outside the fraction:
Now, using our expansion, we can write the top as:
This gives us simply:
So we have something that is easy to integrate. (We would set u = 4x + 1):
OK, great - we have an answer. But didn't I say this was an impossible integral?
Tricky, yes, but impossible?
To perform the above trick, you need to have quite a bit of experience and good observation skills. What if you don't see it?
Let's have another go at substitution (which I abandoned before) and see what happens.
Substitution method
We ignore the 1/2 to make it simpler to follow, and remember to put it back in at the end.
We let u = 4x + 1 and this will give us
du = 4 dx
Now, to re-express the top of the fraction in terms of u, we need to find (x) and (2x + 1) in terms of u.
First, we solve u = 4x + 1 for x
Using this, we find (2x + 1):
So now we have all the pieces we need to integrate our fraction using substitution:
Now, we consider the definitie integral (with our upper and lower limits) and remember to put back the 2 on the bottom):
So we've found an answer. But is it correct? It's the same as we got before, 1/12, but is it correct?
What's going on?
The problem with this question
Now, for those of you who still haven't spotted why this is an impossible question, we need to consider the function and the upper and lower limits given in the question.
If we graph the function we've been trying to integrate, the problem becomes immediately obvious:
In the interval -1 < x < 0, there is a discontinuity!
That is, in this example, we can never let x = -1/4, since this would give 0 on the bottom of the fraction. And of course, we can never divide by 0.
And since we can only integrate continuous functions (ones that have no holes or discontinuities), this integral is impossible!
If the limits of the integral were (say) 0 to 1, there would have been no problem at all, since the function is continuous in that interval.
Conclusion
I have always found it very enlightening to graph the function that I'm differentiating or integrating. It can save you a lot of time wondering why things are going wrong and can help you to make sense of your answer.
And if you are a math teacher, please make sure you give your students questions that actually can be solved! (Unless of course, your aim is to trigger meaningful investigation.)
See the 19 Comments below.
10 Sep 2010 at 11:08 am [Comment permalink]
But it is ok to integrate past a discontinuity (sometimes) - these are considered improper integrals. (With a discontinuity at x=b, you'd need to integrate from the left end to the limit as you approach b from the left, and separately from the limit as you approach b from the right to the right end.)
However, the area in that spike is infinite. So in this particular case the integral does not exist, just as you say.
10 Sep 2010 at 11:54 am [Comment permalink]
Hi Sue and thanks for your input.
Often in math education we are told something is impossible and not to do it, only to find out later there is a whole branch of mathematics that legitimately uses that concept (the square root of negative numbers comes to mind here.)
In this case, I assumed the Scarlett had only recently begun to discover the joys of integration, so I didn't want to complicate things unnecessarily.
But, guilty as charged! I should have included the old "you will learn later that ...".
11 Sep 2010 at 12:07 pm [Comment permalink]
this is an improper integral
11 Sep 2010 at 12:49 pm [Comment permalink]
Hello Tessy. Yes, that is correct.
11 Sep 2010 at 7:25 pm [Comment permalink]
Thanks a million. This is very educative
12 Sep 2010 at 11:07 pm [Comment permalink]
Mr. Murray, you have discussed in a very sequential and logical order that makes this reading very interesting. My students were very impressed with the presentation of the answer. Thanks a lot.
16 Sep 2010 at 11:04 am [Comment permalink]
Its a long time since Ive enjoyed the play of partial differentiation as having taught in junior schools for the last 7 years. But couldnt you see it was discontinuous at x =-1/4 right from the start without graphing ?
16 Sep 2010 at 3:01 pm [Comment permalink]
Hi Barbara. Yes, I could see it was discontinuous (hence my triggers throughout the article, like 'Have any of you spotted why I titled this article "Impossible integration question"?')
But most students won't spot it at first. I felt a graph was the best way to demonstrate the point.
I'm a bit confused, though with your comment about "partial differentiation". That will be a topic in a future article, not this one!
16 Sep 2010 at 3:12 pm [Comment permalink]
Nice article - just wanted to pick up on the partial fractions point. It _is_ possible to express as partial fractions if you set it up as
A + B/(4x+1) + C/(4x+1)^2
The numbers are ugly, of course (I get 3/32, 3/16 and -9/32 - but I haven't had enough sleep to check).
Obviously, that doesn't take away from the main point - but the partial fraction trick is a useful one.
16 Sep 2010 at 4:19 pm [Comment permalink]
Hi Colin. Yes, that does produce an answer. However, A=1/8, B=0 and C=-1/8.
Yah, lack of sleep. It does it to you!
Thanks for the input.
18 Sep 2010 at 12:05 am [Comment permalink]
i never approached functions that are to be integrated from the point of view of graphs.But from now onwards,before attempting to solve any integral,first thing that i am gonna do is to graph the function...
8 Jan 2011 at 1:00 pm [Comment permalink]
can you give some examples with corresponding solutions? especially with logarithmic and exponential examples
thanks............
9 Jan 2011 at 11:12 am [Comment permalink]
Hi Pablu. These 3 chapters should help:
Introduction to integration
Applications of Integrals
Methods of Integration
3 Feb 2011 at 3:36 pm [Comment permalink]
sir,
in your explanation above you used following sentence
"if we reverse the order of the limits, our final answer will be the negative of the answer without reversed limits."
1.if integration gives area under curve,the answer should remain same if we take infinite sum from left to right or right to left right??whats the logic in getting negative answer??i know arithmetic reason behind it i.e we are getting negative answer because we are doing upper limit-lower limit in first case and after reversing,it becomes equivalent to lower limit-upper limit.but how does it make sense in terms of area under curve?
2.we agree that above problem is impossible integral but why are you getting 1/12 as answer??you need to get some condition that says impossibility of above problem mathematically right??why an answer??
5 Feb 2011 at 5:15 pm [Comment permalink]
Hi Gagan. Good questions.
An integral can tell us a lot more than just area under a curve. We can also find out velocity (if we know the function for acceleration) and displacement (if we know velocity) using integration. So there's no problem with getting a negative result in such cases - negative velocity just means speed in the opposite direction and negative displacement just means you measure the distance in the opposite direction.
In this question, the whole point is I can get a NUMBER for my answer, but that number is meaningless. I can see it is meaningless as soon as I look at the graph (or consider the nature of the function if it is simple enough). This is why we should examine our answer carefully and think about what it really means, not just be happy with the fact we get a number.
This is the beginning of mathematical thinking!
7 Apr 2012 at 12:35 pm [Comment permalink]
WONDERFUL EXPLANATION SIR. I THINK THIS IMPOSSIBLE THING GET POSSIBLE DUE TO THE LENGTH OF DISCONTUNITY IS VERY CLOSED TO ZERO. DISCONTINUTY OF FIRST KIND.
THANK YOU SIR.
3 Feb 2015 at 9:40 am [Comment permalink]
Forgive me if I'm wrong, but I have heard that the area between two lines that are going the same direction along the same asymptote is finite. Thereby, the problem IS a problem with a definite integral from -1 to 0.
6 Feb 2015 at 10:44 am [Comment permalink]
Umm, no. In this case, the integral does not exist at the discontinuity.
I think you meant "the area between two CURVES that are going the same direction along the same asymptote is finite". However, as one example, the integral of 1/x^2 from -1 to 1 contradicts that assertion.
3 Jul 2018 at 8:41 pm [Comment permalink]
that solution to murray's problem was great
how can i integrate this problem am stuck which method to use
[x.expX]/[ROOT OF 1-X squared from [1,-1]