Higher Maths (P1) Question 8(c)

less is more

Ok there seems to be some debate as to the correct answer, so a simple post with your answer should help clear things up. BTW it was the Integration shaded area question.

I got 19 units squared

Peslo

less is more wrote: »

Ok there seems to be some debate as to the correct answer, so a simple post with your answer should help clear things up. BTW it was the Integration shaded area question.

I got 19 units squared

Did anybody do what i did, and find the points of intersection, Then find the area enclosed by the curve and the line y=-5 and take away the area enclosed by the line and the line y=-5.
The intervals where from -3 to 3.

I realised at the end that there was another little bit to do, but only realised at the very end!!

NanoZoom555

i got 99/8 units ^2 ,dont think its right tho,im hearing a lot of very different answers

cian1500ww

14

Blackangel

35/3 but i think its very wrong

tramoredude

I knew how to do it, but it looked so time consuming I just went onto Algebra, A girl in my class says she did it in "a minute or two" so she more than likely got the wrong answer too...

I bet it would take 20 minutes to do it out properly! You had to subtract the area of the rectangle + the small triangle in the 4th quadrant too, anyone spot that?

But over all it went well and I did a lot better than I thought I would

life_is_music

yeah i did the rectangle/triangle thingy...

Took me 4 pages to get it out

I got 79/6 i think...... or maybe 77/6

alan4cult

Answer is 32/3 methinks

Peslo

Loads of people i know spent ages doin it, a few said it took 3 or 4 pages to do too

Nanaki

4/3 seems vveeeerrrry wrong!
I got my limits (by plugging y=4-x^2 into the line, I think it was -1 and 3)
Integrated the curve with the limits.
Then did the line (realised halfway through - how the hell do ya intergrate y?!) and subtracted it from the curve.
Kinda noticed that part of the area is below the x axis, does that have a bearing?

Anyone want to post a worked solution?

Cokehead Mother

I'll LaTeX up a pretty solution - give me 5 minutes.

Rich89

Me and another guy i was talking to got 32/3

Nanaki

Cokehead Mother wrote: »

I'll LaTeX up a pretty solution - give me 5 minutes.

excellent, thanks!

TheAmateur

Rich89 wrote: »

Me and another guy i was talking to got 32/3

If that means 10 and two thirds I think I got that too!

jennyq

Also 32/3, did it again just there to make sure.

Decerto

i just did it there and got 32/3 but in exam i used one wrong limit for very last bit under x axis in getting curve instead of 3 and 2 i used 3 and 1/2 anyone know what ill be docked for tht

Cokehead Mother

Turns out I lost the ability to LaTeX efficiently and correctly. As it transpires I have a (horrible) scanner. It's cut off at the end but the answer I have there is 32/3.

orangetictac

Yup thats wat i got!

hippety-pippety

i got 9.... any takers?
it took me a gud while bt 9 is a nice round number so i stuck with it rather than try agen.

shockingpinkdna

I got 8. But then again I can't do Integration!

delta214

I got 32/3 also.
I just moved my x axis down so that the point where the curve and line meet is on the axis. Made it much simpler i thought

alan4cult

delta214 wrote: »

I got 32/3 also.
I just moved my x axis down so that the point where the curve and line meet is on the axis. Made it much simpler i thought

How do you do that method?

littlneutralone

If you do it right but make stupid mental calculator mistakes is it only 3 marks you lose?

Can't recall my answer. I think I did it the same way as posted above but 11.58am leaves little time for prudence.

Cokehead Mother

littlneutralone wrote: »

If you do it right but make stupid mental calculator mistakes is it only 3 marks you lose?

Can't recall my answer. I think I did it the same way as posted above but 11.58am leaves little time for prudence.

1 + 2 = 4 is a slip and you only lose 1 mark for them.

conor563

I got 32/3. It took me ages! I got the area between the curve and the x-axis and the area between the line and the x-axis on both sides. Took the curve area from the positive area of the line and x-axis, then added that answer to the negative answer of the line and x-axis. Sorry if it sounds confusing, it very much is. Haven't a clue if its not the right answer!

irish_boy90

i got 6, looked up the method in my less stress more success.
I got it wrong.

wasn't too far off it though.
didn't bother working out the real answer. P1 is over so I started on P2
only looking to pass H irish and well I did a good oral. paper 2 is geting some much needed attention this weekend

mrskinner

if you want to move the 2 graphs up you add 5 to each,,,,

y = 9 - x^2
and
y = 6 - 2x

integrate the curve from -1 to 3 gives 26.666
the area of the triangle is 12

the difference is 14.6666 which is the shady area?

dyle123

Glad to see many other people got 32/3 because that's what I got and so did another girl in my year!

littlneutralone

Cokehead Mother wrote: »

1 + 2 = 4 is a slip and you only lose 1 mark for them.

Sweet. Hope they follow my method, it looked like a calculator regurgitated on the page. Actually if I'm hoping for stuff, I hope I got the answer.

Anyway. Thanks.:pac:

louise-x

i can't even remember what i got, but looking at all this, it was wrong

Davidius

That question looks big and scary to my poor 5th year brain.

Yuugib

61/6 the second time i v done it

raah!

Does anyone know if it's acceptable to integrate between the graph and the y axis to get the scond part? or are you only able to do that with sidewaysy graphs?

looly

Ah crap....i got 44/5........ah well! i think i did a few bits right so il get my attempt marks anyway lol

Michael Collins

The correct answer is 32/3. The method is quite simple (but might take a while to see how it works):

Area = Int(4 -x^2 - (1 - 2x))dx from -1 up to 3

= 32/3 square units (or 10 and 2/3)

tramoredude

Michael Collins wrote: »

The correct answer is 32/3. The method is quite simple (but might take a while to see how it works):

Area = Int(4 -x^2 - (1 - 2x))dx from -1 up to 3

= 32/3 square units (or 10 and 2/3)

It can't be that easy!

Will you explain it fully please? Its killing me inside!

Michael Collins

tramoredude wrote: »

It can't be that easy!

I'm afraid it looks like it is

tramoredude wrote: »

Will you explain it fully please?

It's not the easiest to explain without diagrams but qualitatively you're looking for the area between the curve and the line. If you subtract the height of the line from the height curve, you've now got a new curve, the area between it and the x-axis is what you're looking for - hence just a simple integration between the limits.

tramoredude wrote: »

Its killing me inside!

That's why I'm not a big fan of exam post-mortems, just concentrate on the next exam, that's the most important one now...

yenoolnairb

I didn't know what to do cos the area went under the x axis and behind the y axis. Integration gets the area between what you're integrating and the x axis (yes?) so wouldn't get the underneath bit i figured so i shifted the whole thing into the positive quadrant by changing xs to x+1s and ys to y+5s and then just integrated between 4 and 0. how does that sound? is it allowed?

~Candy~

Davidius wrote: »

That question looks big and scary to my poor 5th year brain.

haha!! yea.

~Candy~

i think i got the bit under the curve above the x-axis sumthing lyk , -1 to 3
and minus the triangle and i added the the bit under the curve

lol xpt i kinda points...i got sumthing over 3..
so wrong tho, i hope i cud get sum att. marks..

how do ye do b part 2 like !!

Timans

I got 32/3 as well.

But, I did mine really wrong. I hope I'll get 15/20 for it though.

shanedownfall

I'm pretty sure i got 61/6 for this question. Which doesn't seem to be right... But with a lot of people getting it i presume it's due to a small error...

What i did, and i'm really not sure if it's a correct method; is integrate from the y axis to find the area underneath the x-axis. I isolated x, getting (4-y)^1/2, and integrated this using 0 and -5 as limits. I then took away the area of the triangle, and small rectangle that were also accounted for in the integration part.

Does that seem right?... The Part over the x-axis was just basic integration. And i'm sure most will pick up a lot of marks for getting it right.

Challenged

I just found the solution to Q. 8

http://studentxpress.ie/papers/intsoln2008.pdf

LCFin.yay

ye know that the marking schemes are posting on www.examinations.ie soon after the exams over yeah? lol

devinejay

32/3, and so did a friend who I trust (Read;she's a maths nerd)

Michael Collins

Challenged wrote: »

I just found the solution to Q. 8

http://studentxpress.ie/papers/intsoln2008.pdf

Nicely presented solution there. The solution is a bit longer than necessary, however.

LCFin.yay wrote: »

ye know that the marking schemes are posting on www.examinations.ie soon after the exams over yeah? lol

They won't be up until after the results are out, as the marking scheme is decided only after the papers are corrected.

Mr. Blah

I did what the solution shows (with the dividing it up and then adding it at the end) but I got 13. Must of made a slip up somewhere. Least I'll get most marks I suppose :P

Timans

I got the right answer but didn't do it the right way.

How many marks are going for the answer?

less is more

Ok so the general consensus is 32/3.

So unfortunately for me, my nice whole 19 units^2 was WRONG. Thanks guys for raining on my parade lol. I really thought my whole number answer was right... ah well I've never got an integration-area question right before, so why break tradition for the exam? Now onto paper two...

MathsManiac

Why is nobody listening to Michael Collins' words of wisdom? Are you all Dev fans or something?

Surely everyone knows by now, (since it was pointed out in another thread before the exam,) that if f(x) lies above g(x) between x=a and x=b, then, no matter where the axes lie, the area between them is the integral from a to b of [f(x)-g(x)] dx.

So yes, it is as easy as MC said.

starkinter

Unfortunately, this method isn't how most people were taught to approach questions.

Looking at it now, it does make lots of sense.