Shaun F. GCSE Maths tutor, A Level Maths tutor

Shaun F.

£18 - £22 /hr

Currently unavailable: for regular students

Studying: Mathematics with International Study (Masters) - Exeter University

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About me

Hi, I'm Shaun! I'm currently studying for a masters in Mathematics at the University of Exeter. The Sessions I believe that it is very important in mathematics to understand what you are taught, not just blindly learn formulae! As such, I focus on teaching the ideas behind the method, not just the method itself.  I also think it's very important to practise questions yourself, so the sessions are interactive.  The actual structure of the sessions is up to you! I can go through exam papers or focus on particular topics you struggle with. Now What? If you have any questions, feel free to contact me. You can also arrange a free 'Meet the Tutor' session through this website to get to know me a little before you decide whether to go ahead with some sessions.Hi, I'm Shaun! I'm currently studying for a masters in Mathematics at the University of Exeter. The Sessions I believe that it is very important in mathematics to understand what you are taught, not just blindly learn formulae! As such, I focus on teaching the ideas behind the method, not just the method itself.  I also think it's very important to practise questions yourself, so the sessions are interactive.  The actual structure of the sessions is up to you! I can go through exam papers or focus on particular topics you struggle with. Now What? If you have any questions, feel free to contact me. You can also arrange a free 'Meet the Tutor' session through this website to get to know me a little before you decide whether to go ahead with some sessions.

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A
PhysicsA-level (A2)B
ITA-level (A2)B

General Availability

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Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£20 /hr
MathsA Level£20 /hr
MathsGCSE£18 /hr

Questions Shaun has answered

What is the integral of x sin(x) dx?

Find the following integral: ∫ x sin(x) dx

This question is a good candidate for the integration by parts method, as it is the product of two different 'parts'.

Recall that if you have an integral of the form

∫ u(dv/dx) dx

it can be written as

uv – ∫ v(du/dx) dx.

We need to decide which part we will differentiate (as in, which part is u), and which part we will integrate (as in, which part is dv/dx). 

We can note that continuously differentiating sin(x) results in a loop of cos(x), –sin(x), –cos(x), sin(x)..., whereas differentiating x once gives 1.

From this, it seems to make sense that we would want to differentiate the x part (so u is x) and therefore integrate the sin(x) part (so dv/dx is sin(x) ). So, let

u = x, which implies du/dx = 1

and let

dv/dx = sin(x). Integrating this to get v gives v = –cos(x).

So our integral is now of the form required for integration by parts.

∫ x sin(x) dx 

=  ∫ u(dv/dx) dx

= uv –  ∫ v(du/dx) dx

= –x cos(x) –  ∫ –cos(x)*1 dx

= –x cos(x) –  ∫ –cos(x) dx

= –x cos(x) +  ∫ cos(x) dx

The integral of cos(x) is equal to sin(x). We can check this by differentiating sin(x), which does indeed give cos(x). Finally, as with all integration without limits, there must be a constant added, which I'll call c. So the final answer is 

∫ x sin(x) dx = –x cos(x) +  sin(x) + c

Find the following integral: ∫ x sin(x) dx

This question is a good candidate for the integration by parts method, as it is the product of two different 'parts'.

Recall that if you have an integral of the form

∫ u(dv/dx) dx

it can be written as

uv – ∫ v(du/dx) dx.

We need to decide which part we will differentiate (as in, which part is u), and which part we will integrate (as in, which part is dv/dx). 

We can note that continuously differentiating sin(x) results in a loop of cos(x), –sin(x), –cos(x), sin(x)..., whereas differentiating x once gives 1.

From this, it seems to make sense that we would want to differentiate the x part (so u is x) and therefore integrate the sin(x) part (so dv/dx is sin(x) ). So, let

u = x, which implies du/dx = 1

and let

dv/dx = sin(x). Integrating this to get v gives v = –cos(x).

So our integral is now of the form required for integration by parts.

∫ x sin(x) dx 

=  ∫ u(dv/dx) dx

= uv –  ∫ v(du/dx) dx

= –x cos(x) –  ∫ –cos(x)*1 dx

= –x cos(x) –  ∫ –cos(x) dx

= –x cos(x) +  ∫ cos(x) dx

The integral of cos(x) is equal to sin(x). We can check this by differentiating sin(x), which does indeed give cos(x). Finally, as with all integration without limits, there must be a constant added, which I'll call c. So the final answer is 

∫ x sin(x) dx = –x cos(x) +  sin(x) + c

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2 years ago

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