Patrick K. A Level Maths tutor, Mentoring Maths tutor, GCSE Maths tutor

Patrick K.

£18 - £22 /hr

Currently unavailable: for new students

Studying: Mathematics (Masters) - Edinburgh University

5.0
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4 reviews| 8 completed tutorials

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About me

I am a student of mathematics at the University of Edinburgh. I'm passionate about my subject and want to help you discover a love for maths too!

In my sessions we'll use extensive examples to explore the benefits and drawbacks of different mathematical methods, and to highlight areas where care should be taken. We won't just focus on calculation though - I'll try to always emphasise the origins of every formula we use, as this is what makes formulae easy to memorise. Learning will always be guided by you and I'll always take the time to first understand where you struggle to tailor the session to your needs: because everyone learns differently.

I am a student of mathematics at the University of Edinburgh. I'm passionate about my subject and want to help you discover a love for maths too!

In my sessions we'll use extensive examples to explore the benefits and drawbacks of different mathematical methods, and to highlight areas where care should be taken. We won't just focus on calculation though - I'll try to always emphasise the origins of every formula we use, as this is what makes formulae easy to memorise. Learning will always be guided by you and I'll always take the time to first understand where you struggle to tailor the session to your needs: because everyone learns differently.

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Ratings & Reviews

5from 4 customer reviews
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Justine (Parent)

August 10 2016

great tutor

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Justine (Parent)

May 18 2016

Great lesson.

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Josh (Student)

May 17 2016

great helped a lot with my maths

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Justine (Parent)

August 8 2016

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Qualifications

SubjectQualificationGrade
MathematicsScottish highers / Advanced highers (Advanced Higher)A
EnglishScottish highers / Advanced highers (Advanced Higher)A
MusicScottish highers / Advanced highers (Advanced Higher)A
HistoryScottish highers / Advanced highers (Higher)A
PhysicsScottish highers / Advanced highers (Higher)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 Patrick has answered

How do you integrate by parts?

To integrate the product of two functions - say f(x) = xsinx, the product of g(x) = x and h(x) = sin(x) - the we use integration by parts. 

Since integration is the inverse of differentiation, we'll appeal to the Product Rule for differentiation to find a rule for integration. The Product Rule says that (fg)' = f'g + g'f. Rearranging giives:

f'g = (fg)' - g'f.

Integrating we have:

Integral(f'g) = Integral((fg)') - Integral(g'f), or:

Integral(f'g) = fg - Integral(g'f).

This is integration by parts. We designate one of the functions to be f' and one to be g. You can see on the far right we have another integral! For the method to be of use, we need to make the new integral simpler (or at least no more complicated) than the one we began with. So, we want g'f to be simpler then f'g: in other words, when we assign functions f' and g we want g to be the one that gets simpler when differentiated. Let's try an example: let's try and integrate xsinx dx. First, we will choose g to equal x, since x becomes 1 under differentiation, but sinx goes to cosx.

I = Integral(xsinx) = xIntegral(sinx) - Integral(x'Integral(sinx))

I = -xcosx - Integral(-cosx)

I = sinx - xcosx +C

So as long as we choose f' and g with some forethought, it's easy enough to integrate products of functions. Note: Ususally we denote the rule with f' = dv and g = du, so 

Integral u dv  = uv - Integral v du.

This is just a fancy way of stating what we've already established.

To integrate the product of two functions - say f(x) = xsinx, the product of g(x) = x and h(x) = sin(x) - the we use integration by parts. 

Since integration is the inverse of differentiation, we'll appeal to the Product Rule for differentiation to find a rule for integration. The Product Rule says that (fg)' = f'g + g'f. Rearranging giives:

f'g = (fg)' - g'f.

Integrating we have:

Integral(f'g) = Integral((fg)') - Integral(g'f), or:

Integral(f'g) = fg - Integral(g'f).

This is integration by parts. We designate one of the functions to be f' and one to be g. You can see on the far right we have another integral! For the method to be of use, we need to make the new integral simpler (or at least no more complicated) than the one we began with. So, we want g'f to be simpler then f'g: in other words, when we assign functions f' and g we want g to be the one that gets simpler when differentiated. Let's try an example: let's try and integrate xsinx dx. First, we will choose g to equal x, since x becomes 1 under differentiation, but sinx goes to cosx.

I = Integral(xsinx) = xIntegral(sinx) - Integral(x'Integral(sinx))

I = -xcosx - Integral(-cosx)

I = sinx - xcosx +C

So as long as we choose f' and g with some forethought, it's easy enough to integrate products of functions. Note: Ususally we denote the rule with f' = dv and g = du, so 

Integral u dv  = uv - Integral v du.

This is just a fancy way of stating what we've already established.

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2 years ago

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