Differentiating is the process by which when y is given us (as a function of x), we can find dydx.
So, in the case where dydx=xn, working backwards, we get y=1n+1xn+1+C.
Thus, if dydx=4x2, the reverse process gives us y=43x3.
If we begin with a simple case, dydx=x2.
We may write this, if we like, as dy=x2dx.
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 x2dx; and this will not be the same as x2∫dx, because x2 is not a constant. For some of the dx's will be multiplied by big values of x2, and some will be multiplied by small values of x2, 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 xn is “increase the power by one, and divide by the same number as this increased power.” That is to say, x2dx will be changed * to 13x3. Put this into the equation; but don't forget to add the “constant of integration” C at the end. So we get: y=13x3+C.
You have actually performed the integration. How easy!
Let us try another simple case.
Letdydx=ax12, 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 dydx. 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 dy=ax12·dx,∫dy=∫ax12·dx,∫dy=a∫x12dx,y=a×113x13+C.
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 ax12–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.
Let dydx=x2+x3,then dy=x2dx+x3dx.
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: ∫dy=∫(x2+x3)dx=∫x2dx+∫x3dxy=13x3+14x4+C.
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: dydx=xn+b.
This is laughably easy. For you have only to remember that when you differentiated the expression y=ax, the result was dydx=a. Hence, when you work the other way and integrate, the constant reappears multiplied by x. So we get dy=xndx+b·dx,∫dy=∫xndx+∫bdx,y=1n+1xn+1+bx+C.
Here are a lot of examples on which to try your newly acquired powers.
Examples
(1) Given dydx=24x11. Find y.
Ans . y=2x12+C.
(2) Find ∫(a+b)(x+1)dx.
It is (a+b)∫(x+1)dx or (a+b)[∫xdx+∫dx] or (a+b)(x22+x)+C.
(3) Given dudt=gt12. Find u.
Ans . u=23gt32+C.
(4) dydx=x3−x2+x. Find y. dy=(x3−x2+x)dxordy=x3dx−x2dx+xdx;y=∫x3dx−∫x2dx+∫xdx; and y=14x4−13x3+12x2+C.
(5) Integrate 9.75x2.25dx.
Ans . y=3x3.25+C.
All these are easy enough. Let us try another case.
Letdydx=ax−1.
Proceeding as before, we will write dy=ax−1·dx,∫dy=a∫x−1dx.
Well, but what is the integral of x−1dx?
If you look back amongst the results of differentiating x2 and x3 and xn, etc., you will find we never got x−1 from any one of them as the value of dydx. We got 3x2 from x3; we got 2x from x2; we got 1 from x1 (that is, from x itself); but we did not get x−1 from x0, for two very good reasons. First, x0 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−1dx, we see that it does not come in anywhere in the powers of x that are given by the rule: ∫xndx=1n+1xn+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 dydx=x−1 as the result of differentiating the function y=logϵx (see here).
Then, of course, since we know that differentiating logϵx gives us x−1, we know that, by reversing the process, integrating dy=x−1dx will give us y=logϵx. 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=alogϵx+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 logϵx gave x−1, we should have been utterly stuck by the problem how to integrate x−1dx. 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, dydx=a−x2, because a−x2 has never yet been found to result from differentiating anything else.
Another simple case.
Find ∫(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 ∫(x2+3x+2)dx. And this is the same as ∫x2dx+∫3xdx+∫2dx. And performing the integrations, we get 13x3+32x2+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: x−1∫x−1dx=logϵx+C.1x+a∫1x+adx=logϵ(x+a)+C.ϵx∫ϵxdx=ϵx+C.ϵ−x∫ϵ−xdx=−ϵ−x+C (for if y=−1ϵx, dydx=−ϵx×0−1×ϵxϵ2x=ϵ−x). sinx∫sinxdx=−cosx+C.cosx∫cosxdx=sinx+C. Also we may deduce the following: logϵx;∫logϵxdx=x(logϵx−1)+C (for if y=xlogϵx−x, dydx=xx+logϵx−1=logϵx). log10x;∫log10xdx=0.4343x(logϵx−1)+C.ax∫axdx=axlogϵa+C.cosax;∫cosaxdx=1asinax+C (for if y=sinax, dydx=acosax; hence to get cosax one must differentiate y=1asinax). sinax;∫sinaxdx=−1acosax+C.
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, ∬f(x,y,)dxdy 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 x2+y2. Integrating it with respect to x gives us: ∫(x2+y2)dx=13x3+xy2.
Now, integrate this with respect to y: ∫(13x3+xy2)dy=13x3y+13xy3, 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 ∬u·dxdy, 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, volume=∭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 x1 to x2, those for y from y1 to y2, and those for z from z1 to z2, then clearly we have volume=∫z2z1∫y2y1∫x2x1f(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 ∫3x4dx.
(3) Find ∫1ax3dx.
(4) Find ∫(x2+a)dx.
(5) Integrate 5x−72.
(6) Find ∫(4x3+3x2+2x+1)dx.
(7) If dydx=ax2+bx23+cx34; find y.
(8) Find ∫(x2+ax+a)dx.
(9) Find ∫(x+3)3dx.
(10) Find ∫(x+2)(x−a)dx.
(11) Find ∫(√x+3√x)3a2dx.
(12) Find ∫(sinθ−12)dθ3.
(13) Find ∫cos2aθdθ.
(14) Find ∫sin2θdθ.
(15) Find ∫sin2aθdθ.
(16) Find ∫ϵ3xdx.
(17) Find ∫dx1+x.
(18) Find ∫dx1−x.
(1) 4√ax323+C.
(2) −1x3+C.
(3) x44a+C.
(4) 13x3+ax+C.
(5) −2x−52+C.
(6) x4+x3+x2+x+C.
(7) ax24+bx39+cx416+C.
(8) x2+ax+a=x−a+a2+ax+a by division. Therefore the answer is x22−ax+(a2+a)logϵ(x+a)+C. (See pages here and here.)
(9) x44+3x3+272x2+27x+C.
(10) x33+2−a2x2−2ax+C.
(11) a2(2x32+94x43)+C.
(12) −13cosθ−16θ+C.
(13) θ2+sin2aθ4a+C.
(14) θ2−sin2θ4+C.
(15) θ2−sin2aθ4a+C.
(16) 13ϵ3x. % [F1: +C?]
(17) log(1+x)+C.
(18) −logϵ(1−x)+C.