Mary T. A Level Maths tutor, GCSE Maths tutor

Mary T.

£18 - £20 /hr

Currently unavailable: for new students

Studying: Mathematics (Masters) - Durham University

5.0
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3 reviews| 13 completed tutorials

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About me

About me:

I am currently studying maths in my first year at Durham University. Not only do I have a love for my subject but also for teaching it.

In my last year of sixth form I set up a school wide tutoring system which helped many students through their GCSEs and other exams.

Previous experience:

I currently volunteer with the university tutoring a local sixth form student doing his A-level maths. I have also mentored two GCSE students whilst in sixth form.

What to expect:

In a session I would generally aim to help aid understanding of concepts and ideas making it easier for the student to replicate processes in similar questions. I also look partiularly at mark schemes and how examiners may look at an answer as, I believe this is the best way for a student to maximise their potential grade.

My main aim is to ensure that any student is achieving their full potential and making sure that they can become more independent and confident in their mathematical abilities.

If your intersted please feel free to contact me or book a session, I look forward to hearing form you.

About me:

I am currently studying maths in my first year at Durham University. Not only do I have a love for my subject but also for teaching it.

In my last year of sixth form I set up a school wide tutoring system which helped many students through their GCSEs and other exams.

Previous experience:

I currently volunteer with the university tutoring a local sixth form student doing his A-level maths. I have also mentored two GCSE students whilst in sixth form.

What to expect:

In a session I would generally aim to help aid understanding of concepts and ideas making it easier for the student to replicate processes in similar questions. I also look partiularly at mark schemes and how examiners may look at an answer as, I believe this is the best way for a student to maximise their potential grade.

My main aim is to ensure that any student is achieving their full potential and making sure that they can become more independent and confident in their mathematical abilities.

If your intersted please feel free to contact me or book a session, I look forward to hearing form you.

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04/11/2015

Ratings & Reviews

5from 3 customer reviews
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Sonia (Parent)

October 3 2016

Very Helpful

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Sonia (Parent)

October 10 2016

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Lucie (Parent)

April 26 2016

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MatematicsA-level (A2)A*
PhysicsA-level (A2)A*

General Availability

Before 12pm12pm - 5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

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

Questions Mary has answered

∫(1 + 3√x + 5x)dx

For each term the aim is to raise the power of x by 1 and divide by the new power. 

For this question, each part of the expression can be looked at seperately to make things a bit easier:

∫(1 + 3√x + 5x)dx = ∫1dx + ∫3√xdx + ∫5xdx

The first part of the expression can be looked at as 1*x0, so the integral of this is 1*x = x

The second part is a bit more difficult as the power of x isnt a whole number so it can be written as 3*x1/2, the integral of this being     3*x3/2*(2/3) = 2x3/2, (the 2/3 comes from dividing by the new power).

Finally the integral of 5x is easier as the power of x is a whole number and so is easily calculated as 5/2*x2.

Then finally recombining the three part the final answer is:

∫(1 + 3√x + 5x)dx = x + 2x3/2 + (5/2)x+ c

(c is constant and can take any value, this isnt a majorly important part of the question)

For each term the aim is to raise the power of x by 1 and divide by the new power. 

For this question, each part of the expression can be looked at seperately to make things a bit easier:

∫(1 + 3√x + 5x)dx = ∫1dx + ∫3√xdx + ∫5xdx

The first part of the expression can be looked at as 1*x0, so the integral of this is 1*x = x

The second part is a bit more difficult as the power of x isnt a whole number so it can be written as 3*x1/2, the integral of this being     3*x3/2*(2/3) = 2x3/2, (the 2/3 comes from dividing by the new power).

Finally the integral of 5x is easier as the power of x is a whole number and so is easily calculated as 5/2*x2.

Then finally recombining the three part the final answer is:

∫(1 + 3√x + 5x)dx = x + 2x3/2 + (5/2)x+ c

(c is constant and can take any value, this isnt a majorly important part of the question)

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2 years ago

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