question from 2000 paper (1 Viewer)

[timmeh041]

hello all

i have encountered a problem from the 2000 paper ( http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/00mathematics23U.pdf ) that i havn't been able to wrap my head around yet.

Question 7, part c, part ii.

Correct me if i am wrong but when you use integration to find an area, the area is only between the bottom of the curve nad the x axis (or the absolute of the top of the line and x axis)? How then can you simply find the shaded area? After looking up some answers it seem both the BoS marking notes and a seperate book seem to just do a straight integration, minusing the area from the parabole from the straight line, but i fail to see how this could work seeing as the area is both above and under the x axis. I could understand this approach if both line and parabole were pulled up by 2 units so nothing was under the x axis.

Is my understanding completly wrong?
Any help would be much appreciated. This question is killing me.

 

[crono26]

when you have the area between two lines... it doesnt matter where they are in relation to the x axis

so your doing the integral from where they meet from 2 to -1 of the top line (x) - the bottom line (x^2 -2)

not sure i understand ur question

 

[timmeh041]

ok sorry but i think you cleared it up for me.

I was confused about finding the area because it was both above and below the x axis. I thougt the only way to solve it was to divided the area up into a number of sections, find the area of those sections and finally add/subtract them in order to get your answer. I eventually got it, my whole problem was the amount of work allowed for a number of mistakes.

So what you are saying is if you are told to find the area between two lines/functions that span across the top and lower half of a graph, you can just do a start integeration ( integal of function 1 minues function 2) ?

How can you do this when if you had a single line that formed an area underneath the x axis you have to find the absolute of the area? Is it because if you were to move both functions up by y units the area between those two functions is still the same, but if you did the same thing with the single line that was below the x axis the area would be different wither respect to the x axis? ( i think i may have just solved my own problem, but i would like confirmation on this)

I hope you guys understand what i am saying.

 

[forevaunited]

Ok thanks, in trying to explain it to yourself - you explained it to me. I remember i had this problem once, and it took much longer than necessary because i split up the regions + i also made a mistake so got it wrong.

So yeah, since then i have learnt to just subtract the areas from one another, never really knowing why it worked. As you said, because the area would be exactly the same if you shifted it up above the x-axis you can calculate it this way. If you did this with a normal function it would change the area.

Thanks both posters for making me understand it

 

[timmeh041]

haha, glad I have helped you. I am pretty sure that that is why.

 

[Forbidden.]

To find the shaded area between curves with x as the variable.

Image Below

 

[timmeh041]

Thanks Forbidden, I had since blindly accepted the fact after my struggle with the question, steaming mainly from my poor writting/setting out.

Now i understand why it is like that which helps alot.

 

[Queen Llamah]

To find the shaded area between curves with x as the variable.

Image Below

Thats what I thought it was, Untill Timmeh confused the hell outta me. Lol. But atleast i know i'm right. *Is relieved*

 

[timmeh041]

Thats what I thought it was, Untill Timmeh confused the hell outta me. Lol. But atleast i know i'm right. *Is relieved*

If you can remember that then you are fine, but for me to really remember it corectly I NEED to understand how it works. All my rambling up above was just me trying to understand how this method worked, rather then finding out how to blinding solve questions like this.