# What is the mathematical significance of the constant C in an indefinite integral?

As we had seen in an earlier post, calculus bottleneck, calculus presents one of the most difficult topics for the students in higher mathematics. But the problem is not just limited to the students. Teachers feel it too. Too often the emphasis is given on how to solve integration and differentiation problems using “rules” and “methods” while the essence of what is happening is lost. Recently, I asked this question in an interview to a mathematics teacher who was teaching indefinite integration. This teacher had almost a decade of experience in teaching mathematics at +2 level. The teacher tried to answer this question by using an example of the function $$x^{2} + 5$$. Now when we take the derivative of this function, we get

$\dv{ (x^{2} + 5)}{x} = \dv{x^{2}}{x} + \dv{5}{x} = 2x$

as derivative of a constant (5 on our case) is 0). Now the teacher tried to argue, that integration is the reverse of the derivative), so

$\int 2x \, \dd x = \frac{2x^{2}}{2} + C = 2x + C$

After this the teacher tried to argue this $$C$$ represents the constant term (5) in our function $$x^{2} + 5$$. He tried to generalise the result, but he was thinking concretely in terms of the constant in the form of the numbers in the function. The teacher could understand the mechanism of solving the problem, but was not able to explain in clear mathematical terms, why the constant $$C$$ was required in the output of the indefinite integral. This difficulty, I think, partly arose because the teacher only thought in terms solving integrals and derivatives in a particular way, and knew about the connection between the two, but not in a deep way. He did in a way understood the essence of the constant $$C$$, but was not able to understand my question as a general question and hence replied only in terms of concrete functions. Even after repeated probing, the teacher could not get the essence of the question:

why do we add a constant term to the result of the indefinite integral?

To put it in another words, he was not able to generalise an abstract level of understanding from the examples that were discussed. The teacher was thinking only in terms of symbol manipulation rules which are sufficient for problem solving of these types. For example, look at the corresponding rules for differentiation and integration of the function $$x^{n}$$.

$\dv{x^{n}}{x} = n x^{n-1} \iff \int x^{n} \dd x = \frac{x^{n+1}}{n} + C$

Thus, we see according to above correspondence that adding any extra constant $$C$$ to the derivative formula will not affect it. So the teacher claimed it is this constant which appears in the integration rule as well. In a way this is a sort of correct explanation, but it does not get to the mathematical gist of why it is so. What is the core mathematical idea that this constant $$C$$ represents.

Another issue, I think, was the lack of any geometrical interpretation during the discussion. If you ask, what is the geometrical interpretation of the derivative you will get a generic answer along the lines: “It is the tangent to the curve” and for integration the generic answer is along the lines “It is the area under the curve”. Both these answers are correct, but how do these connect to the equivalence above? What is the relationship between the tangent to the curve and area under the curve which allows us to call the integral as the anti-derivative (or is the derivative an anti-integral?). I think to understand these concepts better we have to use the geometrical interpretation of the derivative and the integral from their first definitions.

The basic idea behind the derivative is that it represents the rate of change of a function $$f$$ at a given point. This idea, for an arbitrary function, can be defined (or interpreted) geometrically as:

The derivative of a function $$f$$ at a point $$x_{0}$$ is defined by the slope of the tangent to the graph of the function $$f$$ at the point  $$x = x_{0}$$.

The animation below shows how the slope of the tangent to the sine curve changes. Point $$B$$ in the animation below plots the $$(x, m)$$, where $$m$$ is the slope of the tangent for the given value of $$x$$. Can you mentally trace the locus of point $$B$$? What curve is it tracing?
Now, the tangent to any point on a curve is unique. (Why is it so?) That means if one evaluates a derivative of a function at a point, it will be a unique result for that point.
This being cleared, now let us turn to the indefinite integral. One approach to understanding integration is to consider it as an inverse operation to the derivative, i.e. an anti-derivative.

An anti-derivative is defined as a function $$F(x)$$ whose derivative equals an initial function $$f (x)$$:

$f(x)= \dv{ F(x)}{x}$

Let us take an example of the function $$f(x) = 2x^{2} – 3x$$. The differentiation of this function gives us its derivative $$f'(x) = 4x – 3$$, and its integration gives us anti-derivative.
$F(x) = \frac{2}{3} x^{3} – \frac{3}{2} x^2$

This anti-derivative can be obtained by applying the known rules of differentiation in the reverse order. We can verify that the differentiation of the anti-derivative leads us to the original function.

$F'(x) = \frac{2}{3} 3 x^{2} – \frac{3}{2} 2 x = 2x^{2} – 3x$

Now if add a constant to the function $$F(x)$$, lets say number 4, we get a function $$G(x) = \frac{2}{3} x^{3} – \frac{3}{2} x^2 + 4$$. If we take the derivative of this function $$G(x)$$, we still get our original function back. This is due to the fact that the derivative of a constant is zero. Thus, there can be any arbitrary constant added to the function $$F(x)$$ and it will still be the anti-derivative of the original function $$f(x)$$.

An anti-derivative found for a given function is not unique. If $$F (x)$$ is an anti-derivative (for a function $$f$$ ), then any function $$F(x)+C$$, where $$C$$ is an arbitrary constant, is also an anti-derivative for the initial function because
$\dv{[F(x)+C]}{x} = \dv{ F(x)}{x} + \dv{ C}{x}= \dv{ F(x) }$

But what is the meaning of this constant $$C$$? This means, that each given function $$f (x)$$ corresponds to a family of anti-derivatives, $$F (x) + C$$. The result of adding a constant $$C$$ to any function is that it shifts along the $$Y$$-axis.

Thus what it means for our case of result of the anti-derivative, the resultant would be a family of functions which are separated by $$C$$. For example, let us look at the anti-derivative of $$f (x) = \sin x$$. The curves of anti-derivatives for this function are plotted in will be of the form

$F ( x ) = − \cos x + C$

And this is the reason for adding the arbitrary constant $$C$$ to our result of the anti-derivative: we get a family of curves and the solution is not unique.

Now can we ever know the value of $$C$$? Of course we can, but for this we need to know the some other information about the problem at hand. These can be initial conditions (values) of the variables or the boundary condition. Once we know these we can determine a particular curve (particular solution) from the family of curves for that given problem.

Lev Tarasov – Calculus – Basic Concepts For High Schools (Starts with and explains  the basic mathematical concepts required to understand calculus. The book is in the form of a dialogue between the author and the student, where doubts, misconceptions and aha moments are discussed.)

Morris Kline – Calculus – A physical and intuitive approach (Builds the concepts in the context of the physical problems that calculus was invented to solve.  A book every physics student should read to get an understanding of how mathematics helps solve physical problems.)

Richard  Courant and Fritz John – Introduction to Calculus Analysis (In 2 Volumes) (Standard college level text with in-depth discussions. First volume is rigorous with basic concepts required to conceptually understand the topics and their applications/implications.)

# The Calculus Bottleneck

What if someone told you that learners in high-school don’t actually need calculus as a compulsory subject for a career in STEM? Surely I would disagree. After all, without calculus how will they understand many of the topics in the STEM. For example basic Newtonian mechanics? Another line of thought that might be put forth is that calculus allows learners to develop an interest in mathematics and pursue it as a career. But swell, nothing could be farther from truth. From what I have experienced there are two major categories of students who take calculus in high school. The first category would be students who are just out of wits about calculus, its purpose and meaning. They just see it as another infliction upon them without any significance. They struggle with remembering the formulae and will just barely pass the course (and many times don’t). These students hate mathematics, calculus makes it worse. Integration is opposite of differentiation: but why teach it to us?

The other major category of students is the one who take on calculus but with a caveat. They are the ones who will score in the 80s and 90s in the examination, but they have cracked the exam system per se. And might not have any foundational knowledge of calculus. But someone might ask how can one score 95/100 and still not have foundational knowledge of the subject matter? This is the way to beat the system. These learners are usually drilled in solving problems of a particular type. It is no different than chug and slug. They see a particular problem – they apply a rote learned method to solve it and bingo there is a solution. I have seen students labour “problem sets” — typically hundreds of problems of a given type — to score in the 90s in the papers. This just gives them the ability to solve typical problems which are usually asked in the examinations. Since the examination does not ask for questions based on conceptual knowledge – it never gets tested. Perhaps even their teachers if asked conceptual questions will not be able to handle them — it will be treated like a radioactive waste and thrown out — since it will be out of syllabus.
There is a third minority (a real minority, and may not be real!, this might just be wishful thinking) who will actually understand the meaning and significance of the conceptual knowledge, and they might not score in the 90s. They might take a fancy for the subject due to calculus but the way syllabus is structured it is astonishing that any students have any fascination left for mathematics. Like someone had said: the fascination for mathematics cannot be taught it must be caught. And this is exactly what MAA and NCTM have said in their statement about dropping calculus from high-school.

What the members of the mathematical community—especially those in the Mathematical Association of America (MAA) and the National Council of Teachers of Mathematics (NCTM)—have known for a long time is that the pump that is pushing more students into more advanced mathematics ever earlier is not just ineffective: It is counter-productive. Too many students are moving too fast through preliminary courses so that they can get calculus onto their high school transcripts. The result is that even if they are able to pass high school calculus, they have established an inadequate foundation on which to build the mathematical knowledge required for a STEM career. (emphasis added)

The problem stems from the fact that the foundational topics which are prerequisites for calculus are on shaky grounds. No wonder anything build on top of them is not solid. I remember having very rudimentary calculus in college chemistry, when it was not needed and high-flying into physical meaning of derivatives in physics which was not covered enough earlier. There is a certain mismatch between the expectations from the students and their actual knowledge of the discipline as they come to college from high-school.

Too many students are being accelerated, short-changing their preparation in and knowledge of algebra, geometry, trigonometry, and other precalculus topics. Too many students experience a secondary school calculus course that drills on the techniques and procedures that will enable them to successfully answer standard problems, but are never challenged to encounter and understand the conceptual foundations of calculus. Too many students arrive at college Calculus I and see a course that looks like a review of what they learned the year before. By the time they realize that the expectations of this course are very different from what they had previously experienced, it is often too late to get up to speed.

Though they conclude that with enough solid conceptual background in these prerequisites it might be beneficial for the students to have a calculus course in the highschool.