jennychan12
  • jennychan12
Integral Question. See below,
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
O_o .
jennychan12
  • jennychan12
If \[\int\limits_{0}^{6} (x^2-2x+2)dx\] is approximated by 3 inscribed rectangles of equal width, then the approximation is... 1) 24 B) 26 C) 28 D)76 E) 48
jennychan12
  • jennychan12
sorry typed something wrong. and that 1 is supposed to be an A

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jennychan12
  • jennychan12
so i got the integrated area to be 48, so answers either a b or c
anonymous
  • anonymous
Wait.
anonymous
  • anonymous
Wait, what kind of rectangles? Left, right, or midpoint?
jennychan12
  • jennychan12
the problem doesnt say. i'm assuming left
jennychan12
  • jennychan12
cuz it says inscribed
anonymous
  • anonymous
Well you are right. It's either A, B or C :D . Now can you do it out? :) .
anonymous
  • anonymous
\[\Delta(x)=\frac{ 6-0 }{ 3 }\]
jennychan12
  • jennychan12
i'm confused my friend says to use riemann sum. what's the formula for that?
anonymous
  • anonymous
\[\Delta(x) = 2\]
anonymous
  • anonymous
In this particular example, a Riemann sum is basically the sum of the individual rectangles.
anonymous
  • anonymous
When you learn about definite intergals you will have a good idea about what they actually mean.
jennychan12
  • jennychan12
is it \[\frac{ b-a }{ n } \sum_{i = 1}^{n} f(x) \] ??
anonymous
  • anonymous
No wait, that's right.
anonymous
  • anonymous
So yeah you get the idea. It's the sum of all the rectangles multiples by the change in x,(i.e the width) .
jennychan12
  • jennychan12
that's what i did but i got 28 and the answer's 26
anonymous
  • anonymous
Then you have to use diffrent kinds of rectangles because your answer is correct.
jennychan12
  • jennychan12
26 or 28 ?
anonymous
  • anonymous
It's 26.
anonymous
  • anonymous
But if you use left rectangles then it's 28.
jennychan12
  • jennychan12
why? sorry can u explain?
jennychan12
  • jennychan12
oh.
jennychan12
  • jennychan12
so right hand it's 26 or..?
anonymous
  • anonymous
So this is a very stupid question because it dosen't specify the type of rectangles.
anonymous
  • anonymous
Well right hand is 76 actually.
anonymous
  • anonymous
But it dosen't specify what type of rectangle. Left, right or midpoint.
jennychan12
  • jennychan12
where'd the 26 come from? midpoint?
anonymous
  • anonymous
No, midpoint is 46.
anonymous
  • anonymous
But shall I explain left and right rectangles?
anonymous
  • anonymous
If you use left, you get 28. If you use right you get 76.
anonymous
  • anonymous
Shall I explain?
jennychan12
  • jennychan12
wait so where did the 26 come from. please explain. i'm so lost. :(
anonymous
  • anonymous
26 isn't the correct answer.
jennychan12
  • jennychan12
cuz this other person used 2(f(1)+f(2)+f(4)) and got 26 but i don't understand where the f(1) came from...
jennychan12
  • jennychan12
oh.
anonymous
  • anonymous
Wait hold on.
anonymous
  • anonymous
Yeah, 28 should be correct. I disagree with the answer.
anonymous
  • anonymous
If you use 3 subintervals you CANNOT get 26 in anyway.
anonymous
  • anonymous
The formula would be (2)(f(0)+f(2)+f(4))
jennychan12
  • jennychan12
yeah i did that... and got 28
anonymous
  • anonymous
Which is absolutely correct.
anonymous
  • anonymous
It CANNOT be 26 if you use left rectangles.
anonymous
  • anonymous
if you use right rectangles it's (2)(f(2)+f(4)+f(6)) which is 76.
jennychan12
  • jennychan12
yeah i got 28 and 76 but i chose 28. so my teacher is wrong. hmph.
anonymous
  • anonymous
Yep, you teacher is wrong.
jennychan12
  • jennychan12
thanks for clarifying. :)
anonymous
  • anonymous
In reality the exact value of that integral is 48 but you will learn that later.
jennychan12
  • jennychan12
yeah i know. i calculated it.
anonymous
  • anonymous
Kk.
anonymous
  • anonymous
Good 'ol Riemann's screwing with everybody's head.
anonymous
  • anonymous
I just hope I don't have to do it after my first year in uni :/ . AP calc helped so much though XD .
jennychan12
  • jennychan12
ohhhhhhh i think i got it they used average value
anonymous
  • anonymous
LOL
anonymous
  • anonymous
$$A\simeq2(2^2-2(2)+2+4^2-2(4)+2+6^2-2(6)+2)\\\ \ \ =2(4-4+2+16-8+2+36-12+2)\\\ \ \ =2(38)=76$$ (using right end-points)$$A\simeq2(0^2-2(0)+2+2^2-2(2)+2+4^2-2(4)+2)=2(2+4-4+2+16-8+2)\\\ \ \ =2(14)=28$$ (using left end-points)$$A\simeq2(1^2-2(1)+2+3^2-2(3)+2+5^2-2(5)+2)\\\ \ \ =2(1-2+2+9-6+2+25-20+2)\\\ \ \ =2(13)=26$$ (using midpoints)
anonymous
  • anonymous
^ Reason why they they should specify the question.
jennychan12
  • jennychan12
my friend just assumed that inscribed meant left end points
anonymous
  • anonymous
@oldrin.bataku : Please don't just give the answer next time. The point of this place is to help other people learn. Not give them the answer.
anonymous
  • anonymous
@jennychan12 : That's a flase assumption.
anonymous
  • anonymous
false*
anonymous
  • anonymous
@oldrin.bataku : If you use midpoints, it's 46 not 26.
anonymous
  • anonymous
Lol Oldrin and i have been spending plenty of time explaining problems to different users. I would give him a break xD. I was bored and decided to use this site for the first time today and, oddly enough, I can't stop attempting to solve problems. Probably because I don't want to forget my math whilst im on winter break.
anonymous
  • anonymous
I know but it's just that's it's against the TOS so I try not to. I make that mistake too. It's cool ^_^ .
anonymous
  • anonymous
@Dido525 you're right, 2(5) = 10 so it's actually 2(23) = 46 for the approximate area.
anonymous
  • anonymous
If it requires inscribed rectangles then the approximation must use left end-point... mid- and right- will not be entirely under the parabola.
anonymous
  • anonymous
That's why we have the trapozoidal rule ;) . Ans simpsons rule :P .
anonymous
  • anonymous
and*
anonymous
  • anonymous
But yes @oldrin.bataku is correct. If they MUST be inscribed then you MUSt use left rectangles.
anonymous
  • anonymous
MUST*
anonymous
  • anonymous
@Dido525 if you go into a Scientific Computation course you'll learn about plenty more numerical methods e.g. Runge-Kutta or Adams-Moulton :-p Euler's and Verlet are really common for physics problems.
anonymous
  • anonymous
Conversation is too much for this PoSci major xD.
anonymous
  • anonymous
I looked at Runge-kutta but it makes no sense XD . I am going much deeper into math next year so hopefully then :) .
anonymous
  • anonymous
Sorry that we turned this page into a discussion @jennychan12 .
jennychan12
  • jennychan12
it's aiight.
agent0smith
  • agent0smith
Nobody drew a damn diagram :P
anonymous
  • anonymous
Sigh...
anonymous
  • anonymous
Diagram was drawn in the other thread.
jennychan12
  • jennychan12
^ haha yes
anonymous
  • anonymous
|dw:1355899568667:dw|
anonymous
  • anonymous
Something like that.
agent0smith
  • agent0smith
that's the same thread as this one...
anonymous
  • anonymous
LOL
anonymous
  • anonymous
|dw:1355899651047:dw| That would be incorrect since it does not enclose the curve.
jennychan12
  • jennychan12
yeah i got really confused and tried to do another one to resolve my confusion... :D cuz that other person made me confused
anonymous
  • anonymous
|dw:1355899646424:dw|
anonymous
  • anonymous
|dw:1355899717852:dw|
anonymous
  • anonymous
If you use midpoints it still dosen't enclose the curve fully.
agent0smith
  • agent0smith
Inscribed rectangles *should* mean rectangles within the curve, but it sounds like that didn't work, so the damn question is broken.
anonymous
  • anonymous
Damn question.
anonymous
  • anonymous
It's is 28 though without a doubt.
anonymous
  • anonymous
More than likely your teacher made an error.
jennychan12
  • jennychan12
yeah
agent0smith
  • agent0smith
inscribed rectangles would give 26, but i feel like that was already covered in this mess somewhere
anonymous
  • anonymous
You haven't done "work" problems in your calc class yet have you?
agent0smith
  • agent0smith
Wait, why isnt the answer 26? 3 inscribed rectangles = area of 26.
jennychan12
  • jennychan12
what do u mean? kinetic energy? i do that in physics....
jennychan12
  • jennychan12
@agent0smith can u explain???
anonymous
  • anonymous
there are "work" problems you use calculus for. Pray that you do not have to do them xD
anonymous
  • anonymous
If you use 3 inscribed rectangles it's 28 not 26.
anonymous
  • anonymous
The previous thread came up with 26. this one says 28. let's meet half way and say it's 27...oh wait.
jennychan12
  • jennychan12
lol
agent0smith
  • agent0smith
The height of the first is 1, length is 2 height of the second is 2, length is 2 height of the third is 10, length is 2. 2+4+20
agent0smith
  • agent0smith
How'd you get 28, @Dido525
jennychan12
  • jennychan12
left endpoint is 28
anonymous
  • anonymous
=(2) * ( f(0) + f(2) + f(4) ) = (2) * ( 2 + 2 + 10 ) = 28
agent0smith
  • agent0smith
Inscribed have to be all within the curve, no edges jutting out: it's not f(0), it's f(1), the height of the first rectangle.
agent0smith
  • agent0smith
That is not an inscribed rectangle.
jennychan12
  • jennychan12
that's what that other person did in that other thread 2(f(1)+f(2)+f(4)) = 26 but why would be f(1)??? where does the 1 even come from?
anonymous
  • anonymous
But f(0) is inscribed.
agent0smith
  • agent0smith
I'll draw it jenny. One sec.
anonymous
  • anonymous
I want to see this too.
agent0smith
  • agent0smith
dido, look at the graph: http://www.google.com/search?q=x%5E2-2x%2B2&aq=f&oq=x%5E2-2x%2B2&sugexp=chrome,mod=18&sourceid=chrome&ie=UTF-8 and then look where a rectangle at f(0) would fall... it's not inscribed. it goes over the curve.
anonymous
  • anonymous
Isn't that the same thing as inscribed?
anonymous
  • anonymous
Inscribed meaning within.
agent0smith
  • agent0smith
No, inscribed means it falls completely within the curve, no jutting out over it.
jennychan12
  • jennychan12
okay. i understand that f(1) = 1 but.... oh wait. i see where you're going with this....
anonymous
  • anonymous
|dw:1355900439887:dw|
anonymous
  • anonymous
But rectangles will never cover a curve fully, it's impossible.
anonymous
  • anonymous
Like, it covers it fully but it's an overestimate.
agent0smith
  • agent0smith
anonymous
  • anonymous
Ohh shoot! O_o .
agent0smith
  • agent0smith
Those are inscribed rectangles, they underestimate the area greatly of course.
anonymous
  • anonymous
Ahh, NOW this makes sense.
agent0smith
  • agent0smith
hahah :)
anonymous
  • anonymous
But if it's an underestimate, it's not inscribed is it?
jennychan12
  • jennychan12
ok yeah i see it too.
agent0smith
  • agent0smith
I googled inscribed rectangles just to be sure.
anonymous
  • anonymous
Wait, why did you do a midpoint for the first rectangle?
anonymous
  • anonymous
Instead of a left one?
agent0smith
  • agent0smith
Because that's the only way to have it inscribed. http://en.wikipedia.org/wiki/Inscribed_figure
anonymous
  • anonymous
NOW this all makes sense :D .
agent0smith
  • agent0smith
Inscribed rectangles can be either left or right endpoint
agent0smith
  • agent0smith
or midpoint
agent0smith
  • agent0smith
@jennychan12 you see how it's 26 now? the height of the first is f(1), next is f(2), then f(4)
anonymous
  • anonymous
I would give you my medal if I could :P .
agent0smith
  • agent0smith
haha :) this whole confusion was all over the meaning of the word inscribed :D
anonymous
  • anonymous
Agreed! :D . But hey, this might be n my final so it was good for me too :) .
agent0smith
  • agent0smith
haha good :) hopefully this clears things up Jenny, before you went and told your teacher he/she was wrong because a bunch of people on the internet said so! :D
anonymous
  • anonymous
She is offline :/ .
anonymous
  • anonymous
@jada1701877 : You should have given that to agent.
jennychan12
  • jennychan12
thanks so much! and yeah, i went off at like 11 cuz i went to sleep. :)

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