Mike W. A Level Maths tutor, GCSE Maths tutor, IB Maths tutor, A Leve...

Mike W.

£18 - £25 /hr

Currently unavailable: for regular students

Studying: Mathematics (Masters) - University College London University

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About me

About me

Hi, I'm Mike, I am currently studying maths at UCL in my second year. I have tutored previously whilst in college at GCSE level, and now at university I'm always happy to help with any mathematical aspect of my friends courses. I play for the football team and have enjoyed travelling during the summer.

Tutoring

In the sessions I am happy to cover any part of the subject you require, helping you to understand the process rather than just remember it for the purpose of the exam. I want to make my sessions comfortable and enjoyable, making things easy to understand and you to be happy to interrupt and ask questions as soon as you don't understand anything I have explained. I would like to encourage you to think outside of just the exam; please bring me "what if this...." or  "would this work for..." because this not only makes the subject far more interesting, it is very appealling to universities if you ask questions and think in this manner.

Now what?

Feel free to contact my via any of the methods provided, I'll be happy to talk or even book a free 'Meet the tutor' session for you to meet me and we can discuss anything I have written and if you would like my help!

About me

Hi, I'm Mike, I am currently studying maths at UCL in my second year. I have tutored previously whilst in college at GCSE level, and now at university I'm always happy to help with any mathematical aspect of my friends courses. I play for the football team and have enjoyed travelling during the summer.

Tutoring

In the sessions I am happy to cover any part of the subject you require, helping you to understand the process rather than just remember it for the purpose of the exam. I want to make my sessions comfortable and enjoyable, making things easy to understand and you to be happy to interrupt and ask questions as soon as you don't understand anything I have explained. I would like to encourage you to think outside of just the exam; please bring me "what if this...." or  "would this work for..." because this not only makes the subject far more interesting, it is very appealling to universities if you ask questions and think in this manner.

Now what?

Feel free to contact my via any of the methods provided, I'll be happy to talk or even book a free 'Meet the tutor' session for you to meet me and we can discuss anything I have written and if you would like my help!

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Qualifications

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

General Availability

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Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£20 /hr
MathsA Level£20 /hr
Further MathematicsGCSE£18 /hr
MathsGCSE£18 /hr
PhysicsGCSE£18 /hr
-Personal Statements-Mentoring£22 /hr
.MAT.Uni Admissions Test£25 /hr

Questions Mike has answered

It is given f(x)=(19x-2)/((5-x)(1+6x)) can be expressed A/(5-x)+B/(1+6x) where A and B are integers. i) Find A and B ii) Show the integral of this from 0 to 4 = Kln5

Firstly, we are given that f(x) can be expressed in the above form, so we write this out:

(19x-2)/((5-x)(1+6x)=A/(5-x)+B/(1+6x)

We then multiply by the denominator of f(x):

19x-2=A(1+6x)+B(5-x)

Now we can choose values of x such that each of the brackets equal 0 to find A and B.

x=5  95-2=31A  A=3

x=-1/6  -31/6=(31/6)B  B=-1

So we can write f(x)=3/(5-x)-1/(1+6x)

Now part ii) we can replace the f(x) in the integral with this:

integral(3/(5-x)-1/(1+6x))

We can separate this into

integral(3/(5-x))-integral(1/(1+6x))

Now we want to make the numerator the derivative of the denominator from the form of the answer we're looking for:

-3integral(-1/(5-x))-(1/6)integral(6/(1+6x))

Which equals

-3ln(5-x)-(1/6)ln(1+6x) 

We can sub the limits straight into this:

-3ln1-(1/6)ln25-(-3ln5-(1/6)ln1))

We know ln1=0 so we have

-(1/6)ln25+3ln5

We can rewrite ln25 as 2ln5 to give

(-1/3)ln5+3ln5= (8/3)ln5

i.e. K=8/3

Firstly, we are given that f(x) can be expressed in the above form, so we write this out:

(19x-2)/((5-x)(1+6x)=A/(5-x)+B/(1+6x)

We then multiply by the denominator of f(x):

19x-2=A(1+6x)+B(5-x)

Now we can choose values of x such that each of the brackets equal 0 to find A and B.

x=5  95-2=31A  A=3

x=-1/6  -31/6=(31/6)B  B=-1

So we can write f(x)=3/(5-x)-1/(1+6x)

Now part ii) we can replace the f(x) in the integral with this:

integral(3/(5-x)-1/(1+6x))

We can separate this into

integral(3/(5-x))-integral(1/(1+6x))

Now we want to make the numerator the derivative of the denominator from the form of the answer we're looking for:

-3integral(-1/(5-x))-(1/6)integral(6/(1+6x))

Which equals

-3ln(5-x)-(1/6)ln(1+6x) 

We can sub the limits straight into this:

-3ln1-(1/6)ln25-(-3ln5-(1/6)ln1))

We know ln1=0 so we have

-(1/6)ln25+3ln5

We can rewrite ln25 as 2ln5 to give

(-1/3)ln5+3ln5= (8/3)ln5

i.e. K=8/3

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1 year ago

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