In Section 1, we calculated distances travelled by finding the area under a velocity-time graph. However, we restricted ourselves to ‘straight-line graphs’. Then it’s easy because the areas are all rectangles and triangles (hold this thought…). We promised you we would show you what happens when the lines are not straight – now is the time to do this…
The animation in Section 2 showed you the effect of air resistance on the distance-time graph for a falling object with air resistance. Here is a similar animation for the same object but displaying a velocity-time graph instead (it’s the same motion, just a different graph). Use the slider to alter the amount of air resistance.
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The animation in Section 1 displayed terminal velocity as a constant gradient after a certain point. In contrast, in this animation terminal velocity means constant velocity means a flat line (since the graph is displaying a velocity which is not changing). Just to labour the point, this ‘zero gradient’ means:
(a) zero acceleration, [latex]a=\frac{dv}{dt}=0[/latex],
(b) constant velocity, [latex]v=constant[/latex], or [latex]\frac{dv}{dt}=0[/latex] (if something is constant, then it isn’t changing, so its rate of change is zero)
(c) equal distances travelled in successive equal time intervals (the area under successive equal time intervals are equal, because they are identical rectangles)
The problem we will try to solve on this page is “How far has the object fallen after a certain amount of time?” or conversely “How long will it take to fall a certain distance?”
Look at the pictures below, in which the simplified graph ‘approximates’ (quite badly!) the real situation.
- The first picture is realistic, but has the disadvantage we don’t know how to find the area under it yet.
- The second is easy to calculate the area (triangle + rectangle), but has the disadvantage that it isn’t realistic! If we did find the area it would not be a particularly good approximation to what we wanted to find.
But this is a clue to how to proceed – we can find a technique that gives a better approximation. And perhaps we can refine that technique to give a ‘correct’ answer…
Watch the video to find out how this works.
In the animation below you can alter the rectangle size, and see how it affects the area under the curve. Watch how as the number of rectangles increases, the area tends towards a ‘limiting value’ (we will return to this in section 4).
Numerical integration and analytical integration
The process we have shown, of finding the area under a graph by splitting it into a large number of rectangles of suitably small width, and adding the areas of the individual rectangles, is called numerical integration. Specialist software exists in all sorts of scientific fields to calculate the area under curves.
People who know about the trapezium rule can use trapeziums if they want to, but the principle is the same.
If you know the mathematical function of the curve you can often use ‘analytical’ rather than ‘numerical’ integration to find the area under the curve exactly. Note that if you DON’T know the function of the curve, then numerical integration is your only option. If you can’t integrate the function, then you may need to resort to numerical methods.
We will show you how analytical integration works next.
Integration as the inverse operation to differentiation
In Section 1, we said that:
- An instantaneous rate of change of a quantity is the same thing as the gradient of the tangent line to the graph of the quantity
- The inverse operation (which would find the quantity from the rate of change) is the area under the line of the rate of change
In Section 2 we introduced differentiation as the method to find a gradient for a tangent line. In Section 3 we introduced integration as the method to find areas under curves. This gives us an algebraic method to perform integration – you just do the opposite of differentiation!
Here are some examples. Check you can see that in each case the integration ‘undoes’ the effect of the differentiation:
- If [latex]\frac{d}{dx}(\sin x)=\cos x[/latex], then [latex]\int \cos x \;dx=\sin x[/latex] (with one caveat we will come onto soon)
- If [latex]\frac{d}{dt}(ax^n)=anx^{n-1}[/latex], then [latex]\int ax^{n-1} \: dx=ax^n[/latex] (with the same caveat). Another way of saying this is that [latex]\int ax^n\;dx=\frac{ax^{n+1}}{n+1}[/latex]. Take a minute to check you understand this one, because it’s really important. What we are saying is that if differentiating a power of [latex]x[/latex] involves multiplying by the power and then reducing the power by 1, then integrating a power of [latex]x[/latex] involves the opposite, i.e. increasing the power by one and dividing by the new power (remember that inverse operations for compound procedures involves changing the order of the procedure as well as its type)
- If [latex]\frac{d}{dx}(e^x)=e^x[/latex], then [latex]\int e^x \:dx=e^x[/latex], with that caveat again
Now, there will be loads of other functions that you will need to learn particular techniques to integrate. But that’s all they are – techniques for integration. Don’t lose sight of why you are trying to do the integration in the first place, and what it means.
Definite integrals
Definite integrals are those for which the ‘boundary values’ or limits of the area under the curve are specified. Here’s another video…
What happens when the boundary values on the integral are not specified? Then we have what is called, believe it or not, an ‘indefinite integral’, and that’s where that caveat we keep talking about comes in…
What’s the ‘caveat’? The constant of integration
Our constant use of the word ‘caveat’ may well be annoying you by now. Let’s introduce the key idea via an example: here’s a video…
To summarise the video:
- From a function of velocity, you can discover the acceleration
- From a function of acceleration, you can only find the final velocity if you know the initial velocity
There is a mathematical equivalent to this statement. Follow this logic:
- We know that [latex]\frac{d}{dx}(x^2)=2x[/latex]
- We also know that when we differentiate (find the rate of change of…) a constant, let’s call it [latex]c[/latex], we obtain 0 (because constants don’t change). So [latex]\frac{d}{dx}(c)=0[/latex].
- This means that [latex]\frac{d}{dx}(x^2+c)[/latex] is also 0.
We see that [latex]2x[/latex] is the derivative not just of [latex]x^2[/latex], but of [latex]x^2[/latex] plus any constant. So integrating [latex]2x[/latex] will not result in a unique function, but a function [latex]x^2+c[/latex], in which [latex]c[/latex] is called the constant of integration, and can take any value, unless we have further information that constrains it. Now go back and see how this set of (mathematical) bullet points compares to the previous set of (scientific) bullets just above
So in summary:
- Definite integrals appear with limits on the integration sign, and you get an answer as a numerical value (or in terms of the limits if the limits are not numerical values themselves)
- Integrals without boundary values are called indefinite integrals; the answer must be given with a constant of integration (often called [latex]c[/latex])
So now we can write results without mentioning the word ‘caveat’ again; some indefinite integrals are:
- [latex]\int ax^n \:dx=\frac{ax^{n+1}}{n+1}+c[/latex], for which three examples are [latex]\int a \:dx=ax+c[/latex]; [latex]\int x \:dx=\frac{1}{2}x^2+c[/latex]; [latex]\int x^3 \:dx=\frac{1}{3}x^3+c[/latex]
- [latex]\int \cos x \;dx=\sin x+c[/latex]
Etc… And in general, for the function [latex]f(x)[/latex]:
[latex]\displaystyle \int \frac {df(x)}{dx}dx=f(x)+c[/latex]
which means that integrating the derivative of a function gets you back to that function (with a constant of integration).
Integration applied to familiar situations
We quite often find integration in places where we are used to multiplying quantities together. Here are a few examples…
In just the same way that integration often occurs ‘whenever you expect to multiply quantities’, differentiation often occurs ‘whenever you expect to divide quantities’. For example:
You may know of the concept electric current, [latex]I[/latex], as the charge [latex]Q[/latex] per unit time flowing in a time [latex]t[/latex], so that [latex]I=\frac{Q}{t}[/latex]. But this really gives the average current in that time interval. For the instantaneous current at a particular time, we need the instantaneous rate of change, so [latex]I=\frac{dQ}{dt}[/latex].
Related to the example in the video, force [latex]F[/latex] is often calculated as the change of momentum [latex]\Delta p[/latex]per unit time. But again, calculating that for a given time interval as [latex]F=\frac{\Delta p}{\Delta t}[/latex] gives the average force over the time interval Δt. For the instantaneous force, we need the derivative [latex]F=\frac{dp}{dt}[/latex] (which is a decent statement of Newton’s second law).
So this is the end of our core part on calculus (unless you want to explore Section 3b). We have concentrated on what it’s all about, and why it’s necessary. We aren’t pretending that having looked at these few pages you will be expert at doing anything. Some of you might not mind because you are here for the overview anyway, and just out of interest. But if you are looking to learn how to do stuff, then the most important thing we can say is that becoming expert at doing stuff requires effortful practice, and that’s sort of up to you. Chase down some calculus textbooks and get some practice (you can always come back here if you get bogged down in techniques and want a reminder of the overview). If you are still with us for Section 4, you’ll find we won’t teach you how to solve differential equations either, but we will give you some understanding that will help you learn to do just that if you ever need to.
You’ll find the rest of this website runs the same way. You could say we are taking the easy way out by stopping before teaching you to do things. But there’s loads of material out there for that. Our whole grounds for developing this site is that there isn’t that much stuff out there that provides you with the underlying understanding before you get to have to do it. And that’s what we are trying to do…
Link to Section 03b – A bit more on numerical and analytic integration
If you have met integration before, it is still possible that you haven’t spent much time thinking about the relationship between numerical and analytical integration. If so, here’s a page just for you…
Link to Section 04 – Differential equations
This is a logical next step, since (a) most of science involves these! (b) it follows on from differentiation, and (c) you need integration to solve them!
Link to Section 08 – Work and energy
Or you could go here, if you want to brush up on work and energy…
Link to Home Page
And if you don’t like our suggestions, just go back here…