So, I have two functions which let me reiterate
$ f(t) = -2t + 8$
if $0 \leq $ t $ \leq 3 $
$f(t) = 2 $ if $t \gt 3$
Now, what this looks like is a downward slope from y = 8 and x = 0 that stops reducing at about t = 3 and every f(t) becomes 2.
When I solve for the Area of 2 or A(2) I do the following calculations :
$\frac{1}{2}(2)(6)$ The height at t = 2 is exactly 6, so my answer should be 6. I look at the back of the book and find that the answer is 12. I have no idea why the answer is 12.
I then look at A(5) which should be the area of the first piece of the graph. (Which for arguments sake, we're going to say is 12) and then add that to the area of the base and the height since the remainder of the graph is a rectangle we do not need to do the base times height formula.
So, it becomes an interval of [3,6] which is 6-3 = 3 times the height which is 2. This should just be 3*2 based on my understanding, which is 6. So, 12 + 6 is just 18.
I turn to the back of the book and find 21. I'm kind of lost here, please help me.
This may help:
An area is defined between two limit values of $x$.
The area for any range is simply given by high-school calculus:
$$A = \int\limits_{x = x_l}^{x_u} f(x)\ dx$$