Currently unavailable: for regular students
Degree: Physics (Masters) - Southampton University
I am currently studying physics at the University of Southampton. I first developed my keen interest in science and maths when I had a hugely inspirational teacher in secondary school. I would love to be able to spark a similar interest in others.
I have worked in my local library for the last 4 years where part of my role has been assisting members of the public on the computers and helping train new staff in all aspects of the job. Also in my second year of college I was selected to be a mentor to a couple of first year students. This involved a fair amount of tutoring.
I understand that everybody learns differently and that the best teaching method for one student may seem completely unclear to another. I have the patience to explore various methods of explanation to find one that suits you best.
Each session we will cover the topics that you request. As well as doing worked examples and past papers, we will focus a lot on making sure that you understand where the facts and equations you will be using come from. This way you will be able to apply your knowledge in all situations and not just use it to pass exams.
Feel free to book a meet the tutor session if you have any questions.
I hope to hear from you soon!
|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|Physics||A Level||£20 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Akshay (Student) April 6 2016
Integration by parts is a method of integration used when you are attempting to integrate a function which is the product of two functions. If the two products can be expanded there is usually an easier way to integrate them than integration by parts. For example, x2(x - 4) is easier to integrate when expanded to x3 - 4x2.
The general form of the equation for integration by parts is:
∫f(x)g’(x)dx = f(x)g(x) - ∫g(x)f’(x)dx
where f’(x) is the derivative of f(x). It is also commonly seen as:
∫u dv/dx dx = uv - ∫v du/dx dx
where u and v are both function of x.
A good guideline when deciding which function to use as u (or f(x)) is the acronym LIATE:
Logarithmic e.g. ln(x)
Inverse trigonometry e.g. sin-1(x)
Algebraic e.g. x
Trigonometry e.g. sin(x)
Exponential e.g. ex.
Split the integrand (function to be integrated) in to its 2 products.
E.g. ∫xln(x)dx can be split in to x and ln(x).
Decide which function should be u and which should be dv/dx.
E.g. x is algebraic, ln is logarithmic. Logarithmic comes before algebraic in LIATE so u = ln(x) and dv/dx = x.
Find du/dx and v by differentiating and integrating u and dv/dx respectively.
E.g. u = ln(x), du/dx = x-1, dv/dx = x and v = x2/2
Substitute the variables in to the equation for integration by parts.
E.g. ∫xln(x)dx = ln(x)x2/2 - ∫x-1x2/2 dx = ln(x)x2/2 - ∫x/2 dx.
Evaluate the new integral.
E.g ∫xln(x)dx = ln(x)x2/2 - x2/4 + c = x2/4 (2ln(x) - 1) + c where c is a constant of integration.
Step 5 may require you to perform integration by parts again. Also LIATE does not work in every situation. If it does not work, switch the products used for u and dv/dx and try again.see more