Joseph W. A Level Physics tutor, A Level Maths tutor, A Level Chemist...

Joseph W.

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Chemical Physics (Masters) - Edinburgh University

4.5
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2 reviews

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7 completed lessons

About me

Hey I'm Joe, I'm currently studying Chemistry and Physics at the University of Edinburgh. I'm a second year, and the pain of A levels still hasn't really left me! So I'd love to help you out and hopefully take a bit of the weight off your shoulders. I really enjoy explaining things and I've been told I'm quite good at it, so drop me a message and we'll go from there!

Hey I'm Joe, I'm currently studying Chemistry and Physics at the University of Edinburgh. I'm a second year, and the pain of A levels still hasn't really left me! So I'd love to help you out and hopefully take a bit of the weight off your shoulders. I really enjoy explaining things and I've been told I'm quite good at it, so drop me a message and we'll go from there!

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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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Ratings & Reviews

4.5from 2 customer reviews
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Michelle (Parent from Loughton)

May 24 2016

Well paced. Clear explanations.

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Cheng (Parent from Hounslow)

May 18 2016

Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
PhysicsA-level (A2)A*
ChemistryA-level (A2)A
Chemical Physics Year 1Degree (Bachelors)73%

Subjects offered

SubjectQualificationPrices
ChemistryA Level£20 /hr
MathsA Level£20 /hr
PhysicsA Level£20 /hr
ChemistryGCSE£18 /hr
MathsGCSE£18 /hr
PhysicsGCSE£18 /hr

Questions Joseph has answered

What is the integral of ln(x)? Hint: use parts for this integration

We use parts for this integration even though there is only one term! make ln(x)=1ln(x)

How to do integration by parts

If you have a function that you need to integrate that is two functions of x multiplied by eachother, and youve already tried everything else (inspection and substitution) then you're going to have to use integration by parts. 

Intergrating the function §f(x)g(x) ...

we set one of the fuctions to u' and one to v

normally we'll take the hardest part to integrate as v. 

but in this general example we'll just take f(x)=u' and g(x)=v (putting a ' on a fucntion is to say it has been differentiated) 

The general equation then takes the form 

§(u')v dx = uv - §(v')u dx

or §f(x)g(x) = (§f(x)dx)g(x) - §(§f(x)dx)g'(x)dx which just looks like a mess of symbols.

Sometimes youll have to do this more than once to get the answer, but its a very handy tool in integration! 

So our problem is lnx. To make this by parts we split it up into two terms, 1(lnx) 

Therefore u'=1 and v=lnx so u=x and v'=1/x

This makes the integral....

§-use this as integration sign

§ln(x)dx = xln(x) - §x(1/x)dx

          =xln(x) -§(1)dx

      = xln(x) -x + c 

We use parts for this integration even though there is only one term! make ln(x)=1ln(x)

How to do integration by parts

If you have a function that you need to integrate that is two functions of x multiplied by eachother, and youve already tried everything else (inspection and substitution) then you're going to have to use integration by parts. 

Intergrating the function §f(x)g(x) ...

we set one of the fuctions to u' and one to v

normally we'll take the hardest part to integrate as v. 

but in this general example we'll just take f(x)=u' and g(x)=v (putting a ' on a fucntion is to say it has been differentiated) 

The general equation then takes the form 

§(u')v dx = uv - §(v')u dx

or §f(x)g(x) = (§f(x)dx)g(x) - §(§f(x)dx)g'(x)dx which just looks like a mess of symbols.

Sometimes youll have to do this more than once to get the answer, but its a very handy tool in integration! 

So our problem is lnx. To make this by parts we split it up into two terms, 1(lnx) 

Therefore u'=1 and v=lnx so u=x and v'=1/x

This makes the integral....

§-use this as integration sign

§ln(x)dx = xln(x) - §x(1/x)dx

          =xln(x) -§(1)dx

      = xln(x) -x + c 

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2 years ago

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