First of all, when is and when is ? To figure this out, find out when they are equal.
So, at . Hence on the interval .
How about on ? Well, , so on the interval .
So, the correct set-up for this would be:
I'm getting really frustrated after having spent 5 hours on this question. I get a feeling the entire answer sheet itself is off.
Find the area of the region trapped between the 2 given curves and the indicated lines.
f(x) = 3x
g(x) = x^2
x = -1, x = -2
int[3x-x^2] = 3x^2/2 - x^3/3 ----> call this function z(x)
Take z(-2) - z(-1) = results
What I also did was to integrate each function separately and sub in the the range of x and find the difference.
Apparently, the results I obtain does not tally with the answer sheet that says 31/6
I really don't wish to go through the hassle of typing the entire working. Pretty much in a foul mood now and I need a break. I would appreciate if someone could help.
My answer works out to be 3/2.
It just doesn't make sense.
Since x = 0 and x =3 is where the entire curve intersect, then why can't I use x = [0,3] as the range?
graph y=x^2 && y=3x - Wolfram|Alpha=
It just doesn't make sense mathematically to cover the domain outside of 0 and 3 because the area is not enclosed.
I'm solving this, otherwise I'm not sleeping.
link). You are asked to find the region bounded by four functions. . Where does the line intersect the graph ? It intersects it at . Similarly for . And intersects both functions at . Figuring out where the two graphs intersect can help you find "a" bounded region. It is not the only one. There are several bounded regions. Since you need to use all four functions as bounds, the only bounded region with and as bounds has area as I gave in the previous post.
I understand that. This set up I gave should be correct. But it still doesn't tally.
Answer sheet states 31/6.
Is something wrong? It's can't be the case. This is supposed to be a straight forward and simple abstract question.