I'm a masters graduate with more than three years experience tutoring students in maths and physics for GCSE, A-level and IB exams. All of my students have achieved good results and two of them have been with me for the last three years!

I enjoy teaching and believe that students need a friendly environment where they can make mistakes if they are going to learn and gain confidence. I don't believe there is such a thing as a student who is bad at maths. In my experience, students struggle to keep up with their school maths lessons when they are missing a few basic concepts from early on. I focus on catching up quickly on the basics so that the student can start benefiting normally from their school lessons again.

In between my bachelours and masters studies I worked for two years as a coach for a sports charity in London. This has given me plenty of experience relevant to teaching including experience leading whole-class sessions and working with students with special educational needs.

I am currently enjoying a gap year before starting a PhD in September 2015!

I'm a masters graduate with more than three years experience tutoring students in maths and physics for GCSE, A-level and IB exams. All of my students have achieved good results and two of them have been with me for the last three years!

I enjoy teaching and believe that students need a friendly environment where they can make mistakes if they are going to learn and gain confidence. I don't believe there is such a thing as a student who is bad at maths. In my experience, students struggle to keep up with their school maths lessons when they are missing a few basic concepts from early on. I focus on catching up quickly on the basics so that the student can start benefiting normally from their school lessons again.

In between my bachelours and masters studies I worked for two years as a coach for a sports charity in London. This has given me plenty of experience relevant to teaching including experience leading whole-class sessions and working with students with special educational needs.

I am currently enjoying a gap year before starting a PhD in September 2015!

Enhanced DBS Check

13/09/20115from 1 customer review

Jennifer (Student)

February 5 2015

Owen is brilliant!! He makes a good teacher.

Why do I have to add +c when I integrate?

First of all remember that you only need to include the +c if you are integrating without limits (ie you don’t have the two numbers at the top and bottom of the integration symbol).

To understand why we need the +c when integrating without limits, remember that integrating is the inverse of differentiation. This means that if we differentiate expression A and get back expression B, then we could use integration on B if we wanted to go backwards and figure out what A was originally. Here’s an example:

Expression A - x2 + 3x + 4

Differentiate A with respect to x to get:

Expression B - 2x + 3

We differentiated A by differentiating each small part of it separately and adding them together at the end. So x2 differentiated to 2x, 3x differentiated to 3 and 4 differentiated to 0. Adding these together gives us expression B.

But what if we didn’t know what expression A was and we wanted to use B to figure A out? We could look at the 2x and know it must have come from x2 and we could also look at 3 and know that it came from 3x. But how could we figure out that A included a +4 at the end?

The answer is that without having a bit of extra information, we can’t know about the +4. This is because when we differentiate any constant it goes to 0. So any expression that is exactly the same as A but with a different constant would also differentiate to B. This means we can’t know which of these equations was the original. For example the following expressions all differentiate to B:

Expression C - x2 + 3x + 2

Expression D - x2 + 3x + 30

Expression E - x2 + 3x - 5

Expression F - x2 + 3x

You can see that all expressions that differentiate to B start with x2 + 3x and then have a constant added on the end. So when we integrate B we can say that we get x2 + 3x “plus an unknown constant”. The +c is just how we write “plus an unknown constant” in a nice mathematical way.

Great, but how can I work out what c actually is?

To work out c you need a little bit of extra information about the original equation (this extra information is sometimes called the initial conditions or boundary conditions but don’t worry if you haven’t heard those terms before). Without this information, working out c is impossible.

Here’s an example of a question you might see in a textbook that asks you to find c:

f’(x) is the function obtained by differentiating f(x) with respect to x, and has the form

f’(x) = 4x + 5 where f(1) = 10

Work out an expression for f(x) in terms of x.

We integrate f’(x) to get f(x) = 2x2 + 5x + c. We can then use the extra piece of information f(1) = 10 to figure out what c is. To do this, we simply substitute x = 1 into the expression we just worked out for f and compare it to what we know f(1) really is:

10 = f(1) = 2(12) + 5(1) + c

If we simplify we get

10 = 2 + 5 + c

= 7 + c

Which we rearrange to get c = 3.

Now we know what c is! To finish the question we write out the full equation for f with our newly found c substituted in:

f(x) = 2x2 + 5x + 3

Why do I have to add +c when I integrate?

First of all remember that you only need to include the +c if you are integrating without limits (ie you don’t have the two numbers at the top and bottom of the integration symbol).

To understand why we need the +c when integrating without limits, remember that integrating is the inverse of differentiation. This means that if we differentiate expression A and get back expression B, then we could use integration on B if we wanted to go backwards and figure out what A was originally. Here’s an example:

Expression A - x2 + 3x + 4

Differentiate A with respect to x to get:

Expression B - 2x + 3

We differentiated A by differentiating each small part of it separately and adding them together at the end. So x2 differentiated to 2x, 3x differentiated to 3 and 4 differentiated to 0. Adding these together gives us expression B.

But what if we didn’t know what expression A was and we wanted to use B to figure A out? We could look at the 2x and know it must have come from x2 and we could also look at 3 and know that it came from 3x. But how could we figure out that A included a +4 at the end?

The answer is that without having a bit of extra information, we can’t know about the +4. This is because when we differentiate any constant it goes to 0. So any expression that is exactly the same as A but with a different constant would also differentiate to B. This means we can’t know which of these equations was the original. For example the following expressions all differentiate to B:

Expression C - x2 + 3x + 2

Expression D - x2 + 3x + 30

Expression E - x2 + 3x - 5

Expression F - x2 + 3x

You can see that all expressions that differentiate to B start with x2 + 3x and then have a constant added on the end. So when we integrate B we can say that we get x2 + 3x “plus an unknown constant”. The +c is just how we write “plus an unknown constant” in a nice mathematical way.

Great, but how can I work out what c actually is?

To work out c you need a little bit of extra information about the original equation (this extra information is sometimes called the initial conditions or boundary conditions but don’t worry if you haven’t heard those terms before). Without this information, working out c is impossible.

Here’s an example of a question you might see in a textbook that asks you to find c:

f’(x) is the function obtained by differentiating f(x) with respect to x, and has the form

f’(x) = 4x + 5 where f(1) = 10

Work out an expression for f(x) in terms of x.

We integrate f’(x) to get f(x) = 2x2 + 5x + c. We can then use the extra piece of information f(1) = 10 to figure out what c is. To do this, we simply substitute x = 1 into the expression we just worked out for f and compare it to what we know f(1) really is:

10 = f(1) = 2(12) + 5(1) + c

If we simplify we get

10 = 2 + 5 + c

= 7 + c

Which we rearrange to get c = 3.

Now we know what c is! To finish the question we write out the full equation for f with our newly found c substituted in:

f(x) = 2x2 + 5x + 3