ATAR Notes: Forum

Archived Discussion => Mathematics/Science/Technology => 2014 => Exam Discussion => Victoria => Mathematical Methods CAS => Topic started by: chuck981996 on November 05, 2014, 02:28:17 pm

Title: 2014 Mathematical Methods Exam 1 Paper + Solutions
Post by: chuck981996 on November 05, 2014, 02:28:17 pm: Attached is the 2014 Mathematical Methods Exam 1 Paper and my handwritten solutions. They are error free, in my opinion and the opinion of several of my teachers, except I would note that the graph should go through the point (1,2) (sorry!).

Good luck for tomorrow everyone! Remember, when it gets tough: Pen Down. Breathe. Think. :)

-Charlie
Title: Re: 2014 Mathematical Methods Exam 1 Paper + Solutions
Post by: Dreamweaver on November 12, 2014, 12:00:11 am: Hi, can someone explain why its 12- antidiff please?
Title: Re: 2014 Mathematical Methods Exam 1 Paper + Solutions
Post by: Phy124 on November 12, 2014, 02:07:27 am: Firstly, form a square bounded by $-1\leq x \leq 2$ and $0 \leq y \leq 4$ . Now the area enclosed between the line $y=4$ and the function is equal to the area of this square minus the area between the x axis and the function over the interval $-1\leq x \leq 2$ . The area of the square is given by 3 x 4 = 12 so we have:

$A = 12 - \int^2_{-1} f(x) \ dx$

You could otherwise look at by taking the integral of the upper curve (y=4) minus the lower curve (the function f) over the same interval i.e.

$A = \int^2_{-1}(4 - f(x)) \ dx$

Which if you split up straight away would be $\int^2_{-1}4 \ dx - \int^2_{-1}f(x) \ dx = 12 - \int^2_{-1} f(x) \ dx$ , the same as the other method!

If instead you were wondering why we bounded it to $-1\leq x \leq 2$ rather than $-1\leq x \leq 3$ (and as such had 12 rather than 16) it is because the area in the interval $2\leq x \leq 3$ is not bounded as there is no line along $x=3$ .