For that first question, the standard trapezium rule works and you dont have to think about it being negative when x=1/2. Is this just a coincidence give that this is a log graph or does the trapezium rule formula always give the correct estimate no matter if the graph goes +ve/-ve?
I don't know what the 'standard trapezium rule' is. I assume it just has to do with using a negative length if the graph is below the x axis? If so, then yes that would always work
@@rtwodrew2 its the one thats like h/2(y0+2y1+...2yn-1+yn). I guess all that equation is doing anyway is summing up multiple trapeziums and traingles in the first place. I just plugged the corresponding values without thinking about subtracting that little triangle and it worked. Thanks for the info
Yeah so since it's using y values and not lengths any negative area will be automatically accounted for as you'll be using a negative side length to compute it. Never seen that formula before, although I see now that it is in the pearson textbook, I've always just completely ignored it it seems
this video makes this topic so easy, much appreciated 👍🏼
Hi would you be able to make a vid on the formulas you’ll definitely need for tmua that would be really useful thank you !!!!!
I'll have a think
Hi, for the 2019 P1 Q8, why did u choose a quadratic graph? Could you maybe explain how to choose a graph, and your thought process for it. Thank you.
Simply because it was talking about an underestimate, which we had just done in the previous question, using the same graph
For that first question, the standard trapezium rule works and you dont have to think about it being negative when x=1/2. Is this just a coincidence give that this is a log graph or does the trapezium rule formula always give the correct estimate no matter if the graph goes +ve/-ve?
I don't know what the 'standard trapezium rule' is. I assume it just has to do with using a negative length if the graph is below the x axis? If so, then yes that would always work
@@rtwodrew2 its the one thats like h/2(y0+2y1+...2yn-1+yn). I guess all that equation is doing anyway is summing up multiple trapeziums and traingles in the first place. I just plugged the corresponding values without thinking about subtracting that little triangle and it worked. Thanks for the info
Yeah so since it's using y values and not lengths any negative area will be automatically accounted for as you'll be using a negative side length to compute it. Never seen that formula before, although I see now that it is in the pearson textbook, I've always just completely ignored it it seems
its in formula book too lmao!@@rtwodrew2
hi, are we allowed rulers in tmua exams?
I would have thought so