Alex B. IB Maths tutor, GCSE Maths tutor, A Level Maths tutor, 11 Plu...

Alex B.

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Degree: Physics (Masters) - Durham University

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About me

About me:

I’m Alex and I am in my second year studying physics at Durham University. I have a passion for mathematics and how it can be applied to solve problems in physics. Throughout my A levels, other students would often come to me during and outside of lessons, for assistance in understanding difficult concepts in physics or mathematics. Not only did they find me both friendly and approachable, but also that I was able to explain the concept in a different way to the teacher, which often helped them to get past their difficulties. I found this extremely rewarding and I feel it has equipped me with the skills to effectively communicate ideas to your child, to help them achieve their full potential.

Tutorials:

You can book a free 15 minute meet the tutor session. This will enable me to assess any specific areas to be focussed on during tutorials and find out how to teach your child, so they can learn in the best way possible.
 

Subjects offered

SubjectLevelMy prices
Maths A Level £22 /hr
Physics A Level £22 /hr
Maths GCSE £20 /hr
Physics GCSE £20 /hr
Maths 13 Plus £20 /hr
Maths 11 Plus £20 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA*
PhysicsA-LevelA*
ChemistryA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for new students

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Ratings and reviews

5from 35 customer reviews

Martina (Parent) November 17 2016

Great explanation. Very good lesson!!!

Blessings (Student) November 24 2016

good

Martina (Parent) November 24 2016

Very good lesson , good explanation!

Aled (Student) November 16 2016

good as always
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Questions Alex has answered

Find the integral between 4 and 1 of x^(3/2)-1 with respect to x

When we integrate a function we must first raise the power of x in each term by one. The first term therefore becomes x^(5/2). The second term can be thought of as x^0 which we know that any number to the power zero is equal to 1, so the second term becomes x. We must then divide by the new po...

When we integrate a function we must first raise the power of x in each term by one. The first term therefore becomes x^(5/2). The second term can be thought of as x^0 which we know that any number to the power zero is equal to 1, so the second term becomes x. We must then divide by the new power of x. In the first term we have 1 divided by 5/2 which is equal to 2/5 and in the second term we have -1 divided by 1 which is still equal to -1. Because the integral has limits we do not need to include an integration constant. The integrated expression is (2/5)x^(5/2) - x. The next step in solving this problem is to substitute the limits into the equation. What we do here is take away the value of our integrated function at the lower bound from the value of the integrated function at the upper bound. This gives us the calculation ((2/5) x 4^(5/2) - 4) - ((2/5) x 1^(5/2) - 1). This can be evaluated on your calculator to give 9.4.

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5 months ago

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