George D. A Level Further Mathematics  tutor, A Level Maths tutor, A ...

George D.

£18 - £25 /hr

Currently unavailable: until 30/06/2016

Studying: Mathematics (Masters) - Oxford, Exeter College University

| 1 completed tutorial

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About me

Hi, I’m George. I am a first year mathematics student at Exeter College, Oxford. My passion for problem solving is what first drew me to mathematics and teaching. If you are on this website you have a problem and I would love to help you fix it.

I have experience tutoring young people of all ages in the past, from 11+ tutoring for primary school children to helping students with their Further Maths A-Level.

I have worked as a facilitator at educational camps both in the UK and Slovakia. I also have experience as a private tutor of mathematics and physics.

About our tutorials

For me, tutoring is a rewarding activity and a two-way street. Your interests and needs will guide the sessions.

In our free meet the tutor session we will discuss: what topics you are finding difficult and want to work on, how you work and learn best and what time frame you are looking at.

Can you help me apply to Oxbridge?

I went through that process myself, and I know how difficult and stressful it can be, so I am more than happy to guide you through the application process and help make your application as strong as possible.

I hope to meet you soon.

Hi, I’m George. I am a first year mathematics student at Exeter College, Oxford. My passion for problem solving is what first drew me to mathematics and teaching. If you are on this website you have a problem and I would love to help you fix it.

I have experience tutoring young people of all ages in the past, from 11+ tutoring for primary school children to helping students with their Further Maths A-Level.

I have worked as a facilitator at educational camps both in the UK and Slovakia. I also have experience as a private tutor of mathematics and physics.

About our tutorials

For me, tutoring is a rewarding activity and a two-way street. Your interests and needs will guide the sessions.

In our free meet the tutor session we will discuss: what topics you are finding difficult and want to work on, how you work and learn best and what time frame you are looking at.

Can you help me apply to Oxbridge?

I went through that process myself, and I know how difficult and stressful it can be, so I am more than happy to guide you through the application process and help make your application as strong as possible.

I hope to meet you soon.

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Qualifications

SubjectQualificationGrade
Further MathsA-level (A2)A*
MathsA-level (A2)A*
PhysicsA-level (A2)A*
EcomomicsA-level (A2)A*
STEP 1Uni admission test2
MATUni admission test60

General Availability

Before 12pm12pm - 5pmAfter 5pm
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Subjects offered

SubjectQualificationPrices
EconomicsA Level£20 /hr
Further Mathematics A Level£20 /hr
Further Mathematics A Level£20 /hr
MathsA Level£20 /hr
MathsA Level£20 /hr
PhysicsA Level£20 /hr
PhysicsA Level£20 /hr
MathsGCSE£18 /hr
MathsGCSE£18 /hr
PhysicsGCSE£18 /hr
PhysicsGCSE£18 /hr
Maths13 Plus £18 /hr
Maths13 Plus £18 /hr

Questions George has answered

Given that (2x + 11 )/(2x + 1)(x + 3) ≡ A /(2x + 1) + B /(x + 3) , find the values of the constants A and B. Hence show that the integral from 0 to 2 (2x + 11)/ (2x + 1)(x + 3) dx = ln 15.

First starting from the right hand side.

 A /(2x + 1) + B /(x + 3) = A(x+3)+B(2x+1)/(x+3)(2x+1)

Therefore the numerator = (A+2B)x+(3A+B)

Equating this numorator with the Left hand side we are presented with the two simultaneous equations A+2B=2, 3A+B=11 yielding solutions of B=-1, A=4 by elimination of A

 Hence the integral from 0 to 2  (2x + 11)/ (2x + 1)(x + 3) dx =  integral from 0 to 2 of 4/(2x+1) - 1/(x+3) dx

=[2ln(2x+1) - ln(x+3)] from 0 to 2

= [(2ln5-ln5)-(2ln1-ln3)]

=ln(5)-ln(1/3)

=ln(15)

First starting from the right hand side.

 A /(2x + 1) + B /(x + 3) = A(x+3)+B(2x+1)/(x+3)(2x+1)

Therefore the numerator = (A+2B)x+(3A+B)

Equating this numorator with the Left hand side we are presented with the two simultaneous equations A+2B=2, 3A+B=11 yielding solutions of B=-1, A=4 by elimination of A

 Hence the integral from 0 to 2  (2x + 11)/ (2x + 1)(x + 3) dx =  integral from 0 to 2 of 4/(2x+1) - 1/(x+3) dx

=[2ln(2x+1) - ln(x+3)] from 0 to 2

= [(2ln5-ln5)-(2ln1-ln3)]

=ln(5)-ln(1/3)

=ln(15)

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2 years ago

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