This is just a quick reminder. If you find **any** of this confusing, there is a very important trick for making it easier – practice, practice, practice! It doesn’t take years to master this, it takes a few hours every week for a few weeks. You will become more and more familiar with the techniques and learn intuitively to know which technique to use in which situation.

Let’s start with the very basics.

If , what is its derivative? ie. how do we find a function such that:

The answer of course is . We use our normal differentiation rules which you should now be very familiar with. How about if I told you that there was some function whose derivative was – ie. we reversed the question:

What is a function whose gradient at a point is ?

How do you find what is? We are trying to solve the equation

for . The answer to this is . It’s easy to plug it back in and see that indeed this is correct – take the derivative and see that it does give – when we differentiate the constant, , we get zero. We can say that the antiderivative (or indefinite integral) of is and write:

If the derivative of is then the indefinite integral of is . Actually here it’s easy to see where the comes from. The original question was: What is a function whose gradient at a point is ? Well, we can see that if we find some function which satisfies this, then adding a constant to that function which raises or lowers the whole function by that constant, then it won’t change the gradient.

It will always be easier to check whether the answer to an integral question is correct than finding the answer in the first place.

**If you think you have correctly calculated an integral – check it!! Differentiate it and you should get the INTEGRAND (that function being integrated).**

You should be extremely familiar with the basic rules of integration by now. If not, practice over and over again until they are natural.

### Review:Integration by substitution

The trick is to become familiar enough with the examples to know what substitution will make the problem solvable. There are a fair number of tricks to know, which can’t all be taught in class, but you can get them by studying the problem sheets in detail. It takes some practice but you will, over time, understand WHY what works works.

Example:

What is:

What if we substitute :

Let’s substitute this into the original integral:

we then substitute back again for to get:

We can note something important about the original integral:

is the derivative of which appears in the function .

Warning: This is going to get a bit abstract, but work through it line by line and you will understand.

Let’s say that and . Then we can write what we have as:

Check that you agree with the above

But we know from the chain rule that:

Again – make sure that this makes sense.

We can integrate this equation to get:

we are just integrating both sides of the above equation – nothing fishy going on here!

but

The integral of a differential is the differentiated function, plus a constant

which means that:

or alternatively:

We’ve just renamed ie. is the integral (antiderivative) of .

Now let’s go from this abstract example to the concrete example with and .

The antiderivative of is and so in the above example . Now we have:

because the answer will be as we saw above.

If this doesn’t make sense, run through it a few times and see why we’re making these substitutions.

OK, so we just saw a particular example of an indefinite integration by substitution. Here are some more for you to try…

How about a definite integral by substitution:

This is really shorthand for:

Try the substitution (ie. , in our explication above.):

which means that:

We have to be careful that now we will be integrating over and not , so we have to change the limits too. When , and when , . So the new integral is:

Looking of the graphs written in both the variable and the variable they look very different. The point is that the area is still the same. Both the integrand AND the limits change to get the same answer.

Go here for many examples – work through them and you will master this technique!

grant wilsonJuly 20, 2015 at 11:46 amThis made the lecture very clear! this is very helpful, Thank You!

Jonathan ShockJuly 20, 2015 at 1:54 pmGreat to hear Grant, many thanks for the feedback!

Mathemafrica - UCT MAM1000 lecture notes subject listOctober 3, 2015 at 11:03 am[…] MAM1000 lecture notes part 1, part ii – Integration by substitution review […]

Azhar RohimanOctober 15, 2015 at 2:44 pmDear Dr. Shock, in the last example for the definite integral, when you are specifying F'(x) and g(x), you wrote F'(x) = (1/x^3). Was it not supposed to be F'(x) = 1/x ?

Jonathan ShockOctober 16, 2015 at 5:50 amHi Azhar, this is F'(x)=1/x^3 so that F'(g(x))=1/(ln(x))^3 and thus F'(u)du gives us something very simple when we change to the u variable. Does that make sense?

Azhar RohimanOctober 16, 2015 at 7:13 amOk, I get it now. Thank you.

Thabiso MtsweniJuly 28, 2024 at 9:42 pmThank you very much prof.