Integrating as the Reverse of Differentiating

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

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

If we begin with a simple case, \[ \frac{dy}{dx} = x^2. \]

We may write this, if we like, as \[ dy = x^2\, dx. \]

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
$x^2 \int dx$, 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–see here *ante* –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 changed *
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:
\[
y = \tfrac{1}{3} x^3 + C.
\]

You have actually performed the integration. How easy!

Let us try another simple case.

\begin{align*} Let \dfrac{dy}{dx} &= ax^{12}, \end{align*} where $a$ is any constant multiplier. Well, we found when differentiating (see here) 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} · dx,\\ \int dy &= \int ax^{12} · dx,\\ \int dy &= a \int x^{12}\, dx,\\ y &= a × \tfrac{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.

\begin{align*} \mbox{Let } \frac{dy}{dx} &= x^2 + x^3, \\ \mbox{then } 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 on here, we found that when we differentiated the sum of two separate functions, the differential coefficient 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 &= \tfrac{1}{3} x^3 + \tfrac{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.

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 · 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*

(1) Given $\dfrac{dy}{dx} = 24x^{11}$. Find $y$.

*Ans* . $y = 2x^{12} + C$.

(2) Find $\int (a + b)(x + 1)\, dx$.

It is $(a + b) \int (x + 1)\, dx$ or $(a + b) \left[\int x\, dx + \int dx\right]$ or $(a + b) \left(\dfrac{x^2}{2} + x\right) + C$.

(3) Given $\dfrac{du}{dt} = gt^{\frac{1}{2}}$. Find $u$.

*Ans* . $u = \frac{2}{3} gt^{\frac{3}{2}} + C$.

(4) $\dfrac{dy}{dx} = x^3 - x^2 + x$. Find $y$. \begin{align*} dy &= (x^3 - x^2 + x)\, dx\quad\text{or} \\ dy &= x^3\, dx - x^2\, dx + x\, dx;\quad y = \int x^3\, dx - \int x^2\, dx + \int x\, dx; \end{align*} and \begin{align*} y &= \tfrac{1}{4} x^4 - \tfrac{1}{3} x^3 + \tfrac{1}{2} x^2 + C. \end{align*}

(5) Integrate $9.75x^{2.25}\, dx$.

*Ans* . $y = 3x^{3.25} + C$.

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

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

Proceeding as before, we will write \[ dy = a x^{-1} · dx,\quad \int dy = a \int x^{-1}\, dx. \]

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 could not have
a differential coefficient. *Secondly*, even if it could
be differentiated, its differential coefficient (got by
slavishly following the usual rule) would be $0 × x^{-1}$,
and that multiplication by zero gives it zero value!
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:
\[
\int x^n\, dx = \dfrac{1}{n+1} x^{n+1}.
\]
It is an exceptional case.

Well; but try again. Look through all the various differentials 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 $y = \log_\epsilon x$ (see here).

Then, of course, since we know that differentiating $\log_\epsilon x$ gives us $x^{-1}$, we know that, by reversing the process, integrating $dy = x^{-1}\, dx$ will give us $y = \log_\epsilon x$. 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, \[ y = a \log_\epsilon x + C. \]

*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
$\log_\epsilon x$ 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 to-day,
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.*

Find $\int (x + 1)(x + 2)\, dx$.

On looking at the function to be integrated, you remark that it is the product of two different functions of $x$. You could, you think, integrate $(x + 1)\, dx$ by itself, or $(x + 2)\, dx$ by itself. Of course you could. But what to do with a product? None of the differentiations you have learned have yielded you for the differential coefficient a product like this. Failing such, the simplest thing is to multiply up the two functions, and then integrate. This gives us \[ \int (x^2 + 3x + 2)\, dx. \] And this is the same as \[ \int x^2\, dx + \int 3x\, dx + \int 2\, dx. \] And performing the integrations, we get \[ \tfrac{1}{3} x^3 + \tfrac{3}{2} x^2 + 2x + C. \]

Now that we know that integration is the reverse of differentiation, we may at once look up the differential coefficients we already know, and see from what functions they were derived. This gives us the following integrals ready made: \begin{alignat*}{4} &x^{-1} &&\qquad && \int x^{-1}\, dx &&= \log_\epsilon x + C. \\ %\label{intex2} &\frac{1}{x+a} && && \int \frac{1}{x+a}\, dx &&= \log_\epsilon (x+a) + C. \\ &\epsilon^x && && \int \epsilon^x\, dx &&= \epsilon ^x + C. \\ &\epsilon^{-x} &&&& \int \epsilon^{-x}\, dx &&= -\epsilon^{-x} + C \\ \end{alignat*} (for if $y = - \dfrac{1}{\epsilon^x}$, $\dfrac{dy}{dx} = -\dfrac{\epsilon^x × 0 - 1 × \epsilon^x}{\epsilon^{2x}} = \epsilon^{-x}$). \begin{alignat*}{4} &\sin x && && \int \sin x\, dx &&= -\cos x + C. \\ &\cos x && && \int \cos x\, dx &&= \sin x + C. \\ \end{alignat*} Also we may deduce the following: \begin{alignat*}{4} &\log_\epsilon x; &&&& \int\log_\epsilon x\, dx &&= x(\log_\epsilon x - 1) + C \\ \end{alignat*} (for if $y = x \log_\epsilon x - x$, $\dfrac{dy}{dx} = \dfrac{x}{x} + \log_\epsilon x - 1 = \log_\epsilon x$). \begin{alignat*}{4} &\log_{10} x; &&&& \int\log_{10} x\, dx &&= 0.4343x (\log_\epsilon x - 1) + C. \\ &a^x && && \int a^x\, dx &&= \dfrac{a^x}{\log_\epsilon a} + C. \\ % \label{cosax} &\cos ax; &&&& \int\cos ax\, dx &&= \frac{1}{a} \sin ax + C \\ \end{alignat*} (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{alignat*}{4} &\sin ax; &&&& \int\sin ax\, dx &&= -\frac{1}{a} \cos ax + C. \\ \end{alignat*}

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!

In many cases it is necessary to integrate some expression for two or more variables contained in it; and in that case the sign of integration appears more than once. Thus, \[ \iint f(x,y,)\, dx\, dy \] means that some function of the variables $x$ and $y$ has to be integrated for each. It does not matter in which order they are done. Thus, take the function $x^2 + y^2$. Integrating it with respect to $x$ gives us: \[ \int (x^2+y^2)\, dx = \tfrac{1}{3} x^3 + xy^2. \]

Now, integrate this with respect to $y$: \[ \int (\tfrac{1}{3} x^3 + xy^2)\, dy = \tfrac{1}{3} x^3y + \tfrac{1}{3} xy^3, \] to which of course a constant is to be added. If we had reversed the order of the operations, the result would have been the same.

In dealing with areas of surfaces and of solids, we
have often to integrate both for length and breadth,
and thus have integrals of the form
\[
\iint u · dx\, dy,
\]
where $u$ is some property that depends, at each point,
on $x$ and on $y$. This would then be called a *surface-integral*.
It indicates that the value of all such
elements as $u · dx · dy$ (that is to say, of the value of $u$
over a little rectangle $dx$ long and $dy$ broad) has to be
summed up over the whole length and whole breadth.

Similarly in the case of solids, where we deal with
three dimensions. Consider any element of volume,
the small cube whose dimensions are $dx$ $dy$ $dz$. If
the figure of the solid be expressed by the function
$f(x, y, z)$, then the whole solid will have the *volume-integral*,
\[
\text{volume} = \iiint f(x,y,z) · dx · dy · dz.
\]
Naturally, such integrations have to be taken between
appropriate limits (See here for integration between limits.)
in each dimension; and the
integration cannot be performed unless one knows in
what way the boundaries of the surface depend on
$x$, $y$, and $z$. If the limits for $x$ are from $x_1$ to $x_2$,
those for $y$ from $y_1$ to $y_2$, and those for $z$ from $z_1$
to $z_2$, then clearly we have
\[
\text{volume} = \int_{z1}^{z2} \int_{y1}^{y2} \int_{x1}^{x2} f(x,y,z) · dx · dy · dz.
\]

There are of course plenty of complicated and difficult cases; but, in general, it is quite easy to see the significance of the symbols where they are intended to indicate that a certain integration has to be performed over a given surface, or throughout a given solid space.

(2) Find $\int \frac{3}{x^4}\, dx$.

(3) Find $\int \frac{1}{a} x^3\, dx$.

(4) Find $\int (x^2 + a)\, dx$.

(5) Integrate $5x^{-\frac{7}{2}}$.

(6) Find $\int (4x^3 + 3x^2 + 2x + 1)\, dx$.

(7) If $\dfrac{dy}{dx} = \dfrac{ax}{2} + \dfrac{bx^2}{3} + \dfrac{cx^3}{4}$; find $y$.

(8) Find $\int \left(\frac{x^2 + a}{x + a}\right) dx$.

(9) Find $\int (x + 3)^3\, dx$.

(10) Find $\int (x + 2)(x - a)\, dx$.

(11) Find $\int (\sqrt x + \sqrt[3] x) 3a^2\, dx$.

(12) Find $\int (\sin \theta - \tfrac{1}{2})\, \frac{d\theta}{3}$.

(13) Find $\int \cos^2 a \theta\, d\theta$.

(14) Find $\int \sin^2 \theta\, d\theta$.

(15) Find $\int \sin^2 a \theta\, d\theta$.

(16) Find $\int \epsilon^{3x}\, dx$.

(17) Find $\int \dfrac{dx}{1 + x}$.

(18) Find $\int \dfrac{dx}{1 - x}$.

(1) $\dfrac{4\sqrt{a} x^{\frac{3}{2}}}{3} + C$.

(2) $-\dfrac{1}{x^3} + C$.

(3) $\dfrac{x^4}{4a} + C$.

(4) $\tfrac{1}{3} x^3 + ax + C$.

(5) $-2x^{-\frac{5}{2}} + C$.

(6) $x^4 + x^3 + x^2 + x + C$.

(7) $\dfrac{ax^2}{4} + \dfrac{bx^3}{9} + \dfrac{cx^4}{16} + C$.

(8) $\dfrac{x^2 + a}{x + a} = x - a + \dfrac{a^2 + a}{x + a}$ by division. Therefore the answer is $\dfrac{x^2}{2} - ax + (a^2 + a)\log_\epsilon (x + a) + C$. (See pages here and here.)

(9) $\dfrac{x^4}{4} + 3x^3 + \dfrac{27}{2} x^2 + 27x + C$.

(10) $\dfrac{x^3}{3} + \dfrac{2 - a}{2} x^2 - 2ax + C$.

(11) $a^2(2x^{\frac{3}{2}} + \tfrac{9}{4} x^{\frac{4}{3}}) + C$.

(12) $-\tfrac{1}{3} \cos\theta - \tfrac{1}{6} \theta + C$.

(13) $\dfrac{\theta}{2} + \dfrac{\sin 2a\theta}{4a} + C$.

(14) $\dfrac{\theta}{2} - \dfrac{\sin 2\theta}{4} + C$.

(15) $\dfrac{\theta}{2} - \dfrac{\sin 2a\theta}{4a} + C$.

(16) $\tfrac{1}{3} \epsilon^{3x}$. % [F1: +C?]

(17) $\log(1 + x) + C$.

(18) $-\log_\epsilon (1 - x) + C$.

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