Currently unavailable: for regular students
Degree: Mathematics (Bachelors) - Edinburgh University
I'm currently an undergraduate student studying Mathematics at the University of Edinburgh. I have always found mathematics to be the most interesting and thought-provoking, which is why I have chosen to study it at university and also tutor it online.
One of my longtime hobbies is chess, which I have taught for many years now, with my students ranging from 6 year olds to high school students: This experience has helped develop my teaching and tutoring skills, and I believe I would be easy to get along with no matter what age or personality you are.
I believe that in mathematics, the key is to understand the fundamentals, as everything else will follow from it. Thus during the sessions, in addition to working on practice questions, I would like to put an emphasis on helping you understand the procedures and the concepts, so that you may learn to solve not only a particular question, but all similar types of questions in the future.
If there is anything in particular you would need tutoring in, even outside of mathematics (say questions regarding university applications or exam preparation), please don't hesitate to bring be in contact and I will do my best to help you out. Send me a Webmail or book a Meet the Tutor Session, and we'll try to solve your queries together.
Looking forward to meeting you!
|Maths||A Level||£20 /hr|
|Physics||A Level||£20 /hr|
|Mathematics Higher Level||Baccalaureate||6|
|Physics Higher Level||Baccalaureate||6|
|Chemistry Higher Level||Baccalaureate||5|
|Economics Standard Level||Baccalaureate||6|
|English A Language and Literature Standard Level||Baccalaureate||6|
|Finnish A Literature Standard Level||Baccalaureate||6|
Sometimes while integrating, we may come across an expression that is not a polynomial, and thus we cannot use the convenient power rule to integrate. Consider the function y=xcos(x). It is not immediately clear how we should start we this one, however upon further inspection, we may introduce the technique of Integrating by Parts. Essentially we split the function into two parts, say u and v, and then employ a formula which allows us to integrate them together:
∫u·dv = u·v − ∫v·du
Applying this to our function, we obtain
∫x·cos(x) dx = u·v − ∫v·du
Here we note that integrating cos(x) is a lot simpler than integrating x, and differentiating x is also simpler than differentiating cos(x), so it would make sense to set
u = x and dv = cos(x)
This in turn gives us
du = 1 and v = sin(x)
Thus plugging these values back into our original formula, we get
∫x·cos(x)= x·sin(x) - ∫sin(x)·1
So now, all we need to do is integrate sin(x), which is definitely easier than what we started with. Thus, the end product gives us
∫x·cos(x)= x·sin(x) + cos(x) + C
where C of course is the constant of integration.see more