Jrahs - 1998 Trial (1 Viewer)

[sHin]

I managed to do Q8 D) ii. by splitting the areas up into small pieces, but surely, theres an easier way right?

 

[mojako]

integrate them of course ^_^
make some boundaries.
Assuming you've graphed it,
there are 4 intersections between (and including) pi/6 to 3/2pi right?
You add up three integrals:
1. integrate from pi/6 to pi/2 where the two curves intersect (and if the integral turns out to be negative, take the absolute value, as you probably know) **
2. integrate from pi/2 to "something".
what's this something? you can see from the graph that there's some pattern...
the horizontal distance from pi/6 to pi/2 will be the same as the horizontal distance between this "something" and 3/2pi. so this something is 7/6pi
3. integrate from 7/6pi to 3/2pi **

** the integrals in (1) and (3) will each equal to 1/4 square units, the value you get in part (c) of the question.

Oh when I said "negative" I assumed that you integrated the two functions together, like "cos(x) - sin(2x)". If not then just ignore it if it doesn't make sense.

Why did u attach the other pic? do u need help on that too?

 

[sHin]

Ah, so bloody tedious that question was. Yeah, I pretty much did everything the way you said. I just found the question annoying.

Just ignore the other question, I figured it.

 

[mojako]

oh ok.
when u said divided the ares into small pieces, I thought u were using the Simpson's rule or some other approximation techniques.