Integration as the Reverse of Differentiation

Differentiating is the process by which when \(y\) is given us (as a function of \(x\)), we can find \(\dfrac{dy}{dx}\).

Like every other mathematical operation, the process of differentiation may be reversed; thus, if differentiating \(y = x^4\) gives us \(\dfrac{dy}{dx} = 4x^3\); if one begins with \(\dfrac{dy}{dx} = 4x^3\) one would say that reversing the process would yield \(y = x^4\). But here comes in a curious point. We should get \(\dfrac{dy}{dx} = 4x^3\) if we had begun with any of the following: \(x^4\), or \(x^4 + a\), or \(x^4 + c\), or \(x^4\) with any added constant. So it is clear that in working backwards from \(\dfrac{dy}{dx}\) to \(y\), one must make provision for the possibility of there being an added constant, the value of which will be undetermined until ascertained in some other way. So, if differentiating \(x^n\) yields \(nx^{n-1}\), going backwards from \(\dfrac{dy}{dx} = nx^{n-1}\) will give us \(y = x^n + C\); where \(C\) stands for the yet undetermined possible constant.

Clearly, in dealing with powers of \(x\), the rule for working backwards will be: Increase the power by \(1\), then divide by that increased power, and add the undetermined constant.

So, in the case where \[\frac{dy}{dx} = x^n,\] working backwards, we get \[y = \frac{1}{n + 1} x^{n+1} + C.\]

If differentiating the equation \(y = ax^n\) gives us \[\frac{dy}{dx} = anx^{n-1},\] it is a matter of common sense that beginning with \[\frac{dy}{dx} = anx^{n-1},\] and reversing the process, will give us \[y = ax^n.\] So, when we are dealing with a multiplying constant, we must simply put the constant as a multiplier of the result of the integration.

Thus, if \(\dfrac{dy}{dx} = 4x^2\), the reverse process gives us \(y = \frac{4}{3}x^3\).

But this is incomplete. For we must remember that if we had started with \[y = ax^n + C,\] where \(C\) is any constant quantity whatever, we should equally have found \[\frac{dy}{dx} = anx^{n-1}.\]

So, therefore, when we reverse the process we must always remember to add on this undetermined constant, even if we do not yet know what its value will be.

This process, the reverse of differentiating, is called integration; for it consists in finding the value of the whole quantity \(y\) when you are given only an expression for \(dy\) or for \(\dfrac{dy}{dx}\). Hitherto we have as much as possible kept \(dy\) and \(dx\) together as a derivative: henceforth we shall more often have to separate them.

If we begin with a simple case, \[\frac{dy}{dx} = x^2,\] we may write this, if we like, as \[dy = x^2\, dx.\]

Now this is a “differential equation” which informs us that an element of \(y\) is equal to the corresponding element of \(x\) multiplied by \(x^2\). Now, what we want is the integral; therefore, write down with the proper symbol the instructions to integrate both sides, thus: \[\int dy = \int x^2\, dx.\]

[Note as to reading integrals: the above would be read thus:

“Integral dee-wy equals integral eks-squared dee-eks.”]

We haven’t yet integrated: we have only written down instructions to integrate—if we can. Let us try. Plenty of other fools can do it—why not we also? The left-hand side is simplicity itself. The sum of all the bits of \(y\) is the same thing as \(y\) itself. So we may at once put: \[y = \int x^2\, dx.\]

But when we come to the right-hand side of the equation we must remember that what we have got to sum up together is not all the \(dx\)’s, but all such terms as \(x^2\, dx\); and this will not be the same as \(\displaystyle x^2 \int dx\), because \(x^2\) is not a constant. For some of the \(dx\)’s will be multiplied by big values of \(x^2\), and some will be multiplied by small values of \(x^2\), according to what \(x\) happens to be. So we must bethink ourselves as to what we know about this process of integration being the reverse of differentiation. Now, our rule for this reversed process when dealing with \(x^n\) is “increase the power by one, and divide by the same number as this increased power.” That is to say, \(x^2\, dx\) will be changed¹ to \(\frac{1}{3} x^3\). Put this into the equation; but don’t forget to add the “constant of integration” \(C\) at the end. So we get: \[y = \frac{1}{3} x^3 + C.\]

You have actually performed the integration. How easy!

Let us try another simple case.

Let \[\dfrac{dy}{dx} = ax^{12},\] where \(a\) is any constant multiplier. Well, we found when differentiating that any constant factor in the value of \(y\) reappeared unchanged in the value of \(\dfrac{dy}{dx}\). In the reversed process of integrating, it will therefore also reappear in the value of \(y\). So we may go to work as before, thus \[\begin{align} dy &= ax^{12} \cdot dx,\\ \int dy &= \int ax^{12} \cdot dx,\\ \int dy &= a \int x^{12}\, dx,\\ y &= a \times \frac{1}{13} x^{13} + C. \end{align}\]

So that is done. How easy!

We begin to realize now that integrating is a process of finding our way back, as compared with differentiating. If ever, during differentiating, we have found any particular expression—in this example \(ax^{12}\)—we can find our way back to the \(y\) from which it was derived. The contrast between the two processes may be illustrated by the following remark due to a well-known teacher. If a stranger were set down in Trafalgar Square, and told to find his way to Euston Station, he might find the task hopeless. But if he had previously been personally conducted from Euston Station to Trafalgar Square, it would be comparatively easy to him to find his way back to Euston Station.

Integration of the Sum or Difference of Two Functions

Let \[\begin{align} \frac{dy}{dx} &= x^2 + x^3, \end{align}\] then \[\begin{align} dy &= x^2\, dx + x^3\, dx. \end{align}\]

There is no reason why we should not integrate each term separately: for, as may be seen before, we found that when we differentiated the sum of two separate functions, the derivative was simply the sum of the two separate differentiations. So, when we work backwards, integrating, the integration will be simply the sum of the two separate integrations.

Our instructions will then be: \[\begin{align} \int dy &= \int (x^2 + x^3)\, dx \\ &= \int x^2\, dx + \int x^3\, dx \\ y &= \frac{1}{3} x^3 + \frac{1}{4} x^4 + C. \end{align}\]

If either of the terms had been a negative quantity, the corresponding term in the integral would have also been negative. So that differences are as readily dealt with as sums.

How to Deal with Constant Terms

Suppose there is in the expression to be integrated a constant term—such as this: \[\frac{dy}{dx} = x^n + b.\]

This is laughably easy. For you have only to remember that when you differentiated the expression \(y = ax\), the result was \(\dfrac{dy}{dx} = a\). Hence, when you work the other way and integrate, the constant reappears multiplied by \(x\). So we get \[\begin{align} dy &= x^n\, dx + b \cdot dx, \\ \int dy &= \int x^n\, dx + \int b\, dx, \\ y &= \frac{1}{n+1} x^{n+1} + bx + C. \end{align}\]

Here are a lot of examples on which to try your newly acquired powers.

Examples

All these are easy enough. Let us try another case.

Let \[\begin{align} \dfrac{dy}{dx} &= ax^{-1}. \end{align}\]

Proceeding as before, we will write \[\begin{align} dy &= a x^{-1} \cdot dx,\\ \int dy &= a \int x^{-1}\, dx. \end{align}\]

Well, but what is the integral of \(x^{-1}\, dx\)?

If you look back amongst the results of differentiating \(x^2\) and \(x^3\) and \(x^n\), etc., you will find we never got \(x^{-1}\) from any one of them as the value of \(\dfrac{dy}{dx}\). We got \(3x^2\) from \(x^3\); we got \(2x\) from \(x^2\); we got \(1\) from \(x^1\) (that is, from \(x\) itself); but we did not get \(x^{-1}\) from \(x^0\), for two very good reasons. First, \(x^0\) is simply \(= 1\), and is a constant, and the derivative of a constant is zero (not \(x^{-1}\)). Secondly, even if we differentiate it by the Power Rule, its derivative would be \(0 \times x^{-1}\), and that multiplication by zero gives it zero value!² Therefore, when we now come to try to integrate \(x^{-1}\, dx\), we see that it does not come in anywhere in the powers of \(x\) that are given by the rule: \[\int x^n\, dx = \dfrac{1}{n+1} x^{n+1}+C.\] It is an exceptional case.

Well; but try again. Look through all the various derivatives obtained from various functions of \(x\), and try to find amongst them \(x^{-1}\). A sufficient search will show that we actually did get \(\dfrac{dy}{dx} = x^{-1}\) as the result of differentiating the function \(y = \ln x\) (see here).

Then, of course, since we know that differentiating \(\ln x\) gives us \(x^{-1}\), we know that, by reversing the process, integrating \(dy = x^{-1}\, dx\) will give us \(y = \ln x\). But we must not forget the constant factor \(a\) that was given, nor must we omit to add the undetermined constant of integration. This then gives us as the solution to the present problem, \[y = a \ln x + C.\] The above formula is not acceptable when \(x\) is negative as logarithms of negative numbers are imaginary. Now the question is: What function when differentiated gives \(x^{-1}\) when \(x<0\)? When \(x<0\), we can take the logarithm of \(-x\) as \(-x>0\). Let’s see what the derivative of \(\ln(-x)\) is. To differentiate \(\ln(-x)\), let \(u=-x\) and then apply the Chain Rule: \[\begin{align} \frac{d(\ln(-x))}{dx}&=\frac{d(\ln u)}{du}\cdot\frac{du}{dx}\\ &=\frac{1}{u}\times (-1)\\ &=\frac{1}{-x}\times(-1)\\ &=\frac{1}{x} \end{align}\] Hence \[\dfrac{d(\ln(-x))}{dx}=\frac{1}{x} \qquad\text{when }x<0.\] Since differentiating \(\ln(-x)\) gives us \(x^{-1}\) for \(x<0\), integrating \(dy=x^{-1}dx\) for \(x<0\) will give us \(y=\ln(-x)\). Knowing this we can write \[\int \frac{1}{x}dx=\left\{\begin{align} &\ln x+C_1 &&\text{if }x>0\\ &\ln(-x)+C_2 &&\text{if }x<0 \end{align}\right.\] where \(C_1\) and \(C_2\) are two arbitrary constants that do not need to be equal. For any interval not containing \(x=0\), we can combine these two cases and write \[\int\frac{1}{x}dx=\ln |x|+C\] where \(|x|\) is called the absolute value of \(x\) and is defined as \[|x|=\left\{\begin{align} &x &&\text{if }x\geq 0\\ &-x &&\text{if }x<0 \end{align}\right.\] Therefore, \[\int\frac{a}{x}dx=a\ln|x|+C.\]

In summary \[\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\displaystyle \int{x^n dx}=\left\{\begin{align} &\frac{1}{n+1}x^{n+1}+C &&(n\neq -1)\\[9pt] &\ln|x|+C && (n=-1) \end{align}\right.}\]

Note—Here note this very remarkable fact, that we could not have integrated in the above case if we had not happened to know the corresponding differentiation. If no one had found out that differentiating \(\ln x\) gave \(x^{-1}\), we should have been utterly stuck by the problem how to integrate \(x^{-1}\, dx\). Indeed it should be frankly admitted that this is one of the curious features of the integral calculus:—that you can’t integrate anything before the reverse process of differentiating something else has yielded that expression which you want to integrate. No one, even today, is able to find the general integral of the expression, \[\frac{dy}{dx} = a^{-x^2},\] because \(a^{-x^2}\) has never yet been found to result from differentiating anything else.

Another simple case:

Some other Integrals

Now that we know that integration is the reverse of differentiation, we may at once look up the derivatives we already know, and see from what functions they were derived. This gives us the following integrals ready made: \begin{align} &\boldsymbol{y} && &&   \int \boldsymbol{y\, dx}      && \\   \hline\\ &x^{-1}; &&\qquad &&  \int x^{-1}\, dx      &&= \ln |x| + C. \\ % %\label{intex2} &\frac{1}{x+a}; && &&  \int \frac{1}{x+a}\, dx &&= \ln |x+a| + C. \\ % &e^x; && &&  \int e^x\, dx    &&= e ^x + C. \\ % &e^{-x}; &&&&  \int e^{-x}\, dx &&= -e^{-x} + C % \end{align} (for if \(y = - \dfrac{1}{e^x}\),\(\dfrac{dy}{dx} = -\dfrac{e^x \times 0 - 1 \times e^x}{e^{2x}} = e^{-x}\)). \begin{align} &\sin x; &&  &&  \int \sin x\, dx        &&= -\cos x + C. \\ % &\cos x; &&  &&  \int \cos x\, dx        &&= \sin x + C. \\ \end{align} Also we may deduce the following: \[\begin{align} &\ln x; &&&& \int\ln x\, dx &&= x(\ln x - 1) + C \end{align}\] (for if \(y = x \ln x - x\),\(\dfrac{dy}{dx} = \dfrac{x}{x} + \ln x - 1 = \ln x\)). \[\begin{align} &\log_{10} x; &&&& \int\log_{10} x\, dx &&= \frac{1}{\ln 10} x (\ln x - 1) + C. \end{align}\] (for \(\displaystyle \log_{10}x=\frac{\ln x}{\ln 10}\) and \(\displaystyle \int \ln x\, dx=x(\ln x-1)+\text{some constant}\)) \\begin{align} &a^x; &&  &&  \int a^x\, dx        &&= \dfrac{a^x}{\ln a} + C. \\ % % \label{cosax} &\cos ax; &&&& \int\cos ax\, dx     &&= \frac{1}{a} \sin ax + C \end{align} (for if \(y = \sin ax\), \(\dfrac{dy}{dx} = a \cos ax\); hence to get \(\cos ax\) one must differentiate \(y = \dfrac{1}{a} \sin ax\)). \[\begin{align} &\sin ax; &&&& \int\sin ax\, dx &&= -\frac{1}{a} \cos ax + C. \end{align}\]

Try also \(\cos^2\theta\); a little dodge will simplify matters: \[\begin{gathered} \cos 2\theta = \cos^2\theta - \sin^2\theta = {2\cos^2 \theta - 1}; \end{gathered}\] hence \[\begin{gathered} \cos^2\theta = \frac{1}{2}({\cos 2\theta + 1}), \end{gathered}\]

and \[\begin{align} \int\cos^2 \theta\, d\theta &= \frac{1}{2} \int (\cos 2\theta + 1)\, d\theta \\ &= \frac{1}{2} \int \cos 2 \theta\, d\theta + \frac{1}{2} \int d\theta. \\ &= \frac{\sin 2\theta}{4} + \frac{\theta}{2} + C. \end{align}\] (see also here).

See also the Table of Standard Forms. You should make such a table for yourself, putting in it only the general functions which you have successfully differentiated and integrated. See to it that it grows steadily!

Exercises