jennychan12
Integral Question.
See below,
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Dido525
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O_o .
jennychan12
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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
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sorry typed something wrong. and that 1 is supposed to be an A
jennychan12
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so
i got the integrated area to be 48, so answers either a b or c
Dido525
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Wait.
Dido525
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Wait, what kind of rectangles? Left, right, or midpoint?
jennychan12
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the problem doesnt say. i'm assuming left
jennychan12
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cuz it says inscribed
Dido525
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Well you are right. It's either A, B or C :D . Now can you do it out? :) .
Dido525
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\[\Delta(x)=\frac{ 6-0 }{ 3 }\]
jennychan12
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i'm confused
my friend says to use riemann sum.
what's the formula for that?
Dido525
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\[\Delta(x) = 2\]
Dido525
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In this particular example, a Riemann sum is basically the sum of the individual rectangles.
Dido525
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When you learn about definite intergals you will have a good idea about what they actually mean.
jennychan12
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is it \[\frac{ b-a }{ n } \sum_{i = 1}^{n} f(x) \] ??
Dido525
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No wait, that's right.
Dido525
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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
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that's what i did but i got 28
and the answer's 26
Dido525
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Then you have to use diffrent kinds of rectangles because your answer is correct.
jennychan12
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26 or 28 ?
Dido525
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It's 26.
Dido525
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But if you use left rectangles then it's 28.
jennychan12
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why? sorry can u explain?
jennychan12
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oh.
jennychan12
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so right hand it's 26 or..?
Dido525
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So this is a very stupid question because it dosen't specify the type of rectangles.
Dido525
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Well right hand is 76 actually.
Dido525
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But it dosen't specify what type of rectangle. Left, right or midpoint.
jennychan12
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where'd the 26 come from? midpoint?
Dido525
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No, midpoint is 46.
Dido525
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But shall I explain left and right rectangles?
Dido525
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If you use left, you get 28. If you use right you get 76.
Dido525
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Shall I explain?
jennychan12
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wait so where did the 26 come from.
please explain. i'm so lost. :(
Dido525
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26 isn't the correct answer.
jennychan12
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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
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oh.
Dido525
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Wait hold on.
Dido525
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Yeah, 28 should be correct. I disagree with the answer.
Dido525
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If you use 3 subintervals you CANNOT get 26 in anyway.
Dido525
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The formula would be (2)(f(0)+f(2)+f(4))
jennychan12
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yeah i did that...
and got 28
Dido525
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Which is absolutely correct.
Dido525
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It CANNOT be 26 if you use left rectangles.
Dido525
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if you use right rectangles it's (2)(f(2)+f(4)+f(6)) which is 76.
jennychan12
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yeah i got 28 and 76 but i chose 28.
so my teacher is wrong. hmph.
Dido525
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Yep, you teacher is wrong.
jennychan12
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thanks for clarifying. :)
Dido525
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In reality the exact value of that integral is 48 but you will learn that later.
jennychan12
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yeah i know. i calculated it.
Dido525
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Kk.
yun2thejae
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Good 'ol Riemann's screwing with everybody's head.
Dido525
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I just hope I don't have to do it after my first year in uni :/ . AP calc helped so much though XD .
jennychan12
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ohhhhhhh i think i got it
they used average value
yun2thejae
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LOL
oldrin.bataku
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$$A\simeq2(2^2-2(2)+2+4^2-2(4)+2+6^2-2(6)+2)\\\ \ \ =2(4-4+2+16-8+2+36-12+2)\\\ \ \ =2(38)=76$$ (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)
Dido525
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^ Reason why they they should specify the question.
jennychan12
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my friend just assumed that inscribed meant left end points
Dido525
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@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.
Dido525
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@jennychan12 : That's a flase assumption.
Dido525
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false*
Dido525
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@oldrin.bataku : If you use midpoints, it's 46 not 26.
yun2thejae
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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.
Dido525
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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 ^_^ .
oldrin.bataku
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@Dido525 you're right, 2(5) = 10 so it's actually 2(23) = 46 for the approximate area.
oldrin.bataku
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If it requires inscribed rectangles then the approximation must use left end-point... mid- and right- will not be entirely under the parabola.
Dido525
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That's why we have the trapozoidal rule ;) . Ans simpsons rule :P .
Dido525
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and*
Dido525
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But yes @oldrin.bataku is correct. If they MUST be inscribed then you MUSt use left rectangles.
Dido525
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MUST*
oldrin.bataku
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@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.
yun2thejae
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Conversation is too much for this PoSci major xD.
Dido525
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I looked at Runge-kutta but it makes no sense XD . I am going much deeper into math next year so hopefully then :) .
Dido525
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Sorry that we turned this page into a discussion @jennychan12 .
jennychan12
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it's aiight.
agent0smith
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Nobody drew a damn diagram :P
Dido525
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Sigh...
yun2thejae
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Diagram was drawn in the other thread.
jennychan12
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^ haha yes
Dido525
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|dw:1355899568667:dw|
Dido525
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Something like that.
agent0smith
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that's the same thread as this one...
yun2thejae
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LOL
Dido525
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|dw:1355899651047:dw|
That would be incorrect since it does not enclose the curve.
jennychan12
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yeah i got really confused and tried to do another one to resolve my confusion...
:D cuz that other person made me confused
oldrin.bataku
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|dw:1355899646424:dw|
Dido525
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|dw:1355899717852:dw|
Dido525
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If you use midpoints it still dosen't enclose the curve fully.
agent0smith
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Inscribed rectangles *should* mean rectangles within the curve, but it sounds like that didn't work, so the damn question is broken.
Dido525
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Damn question.
Dido525
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It's is 28 though without a doubt.
oldrin.bataku
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More than likely your teacher made an error.
jennychan12
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yeah
agent0smith
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inscribed rectangles would give 26, but i feel like that was already covered in this mess somewhere
yun2thejae
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You haven't done "work" problems in your calc class yet have you?
agent0smith
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Wait, why isnt the answer 26? 3 inscribed rectangles = area of 26.
jennychan12
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what do u mean?
kinetic energy? i do that in physics....
jennychan12
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@agent0smith can u explain???
yun2thejae
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there are "work" problems you use calculus for. Pray that you do not have to do them xD
Dido525
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If you use 3 inscribed rectangles it's 28 not 26.
yun2thejae
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The previous thread came up with 26. this one says 28. let's meet half way and say it's 27...oh wait.
jennychan12
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lol
agent0smith
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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
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How'd you get 28, @Dido525
jennychan12
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left endpoint is 28
Dido525
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=(2) * ( f(0) + f(2) + f(4) )
= (2) * ( 2 + 2 + 10 )
= 28
agent0smith
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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
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That is not an inscribed rectangle.
jennychan12
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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?
Dido525
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But f(0) is inscribed.
agent0smith
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I'll draw it jenny. One sec.
Dido525
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I want to see this too.
Dido525
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Isn't that the same thing as inscribed?
Dido525
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Inscribed meaning within.
agent0smith
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No, inscribed means it falls completely within the curve, no jutting out over it.
jennychan12
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okay. i understand that f(1) = 1 but....
oh wait. i see where you're going with this....
Dido525
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|dw:1355900439887:dw|
Dido525
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But rectangles will never cover a curve fully, it's impossible.
Dido525
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Like, it covers it fully but it's an overestimate.
agent0smith
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Dido525
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Ohh shoot! O_o .
agent0smith
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Those are inscribed rectangles, they underestimate the area greatly of course.
Dido525
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Ahh, NOW this makes sense.
agent0smith
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hahah :)
Dido525
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But if it's an underestimate, it's not inscribed is it?
jennychan12
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ok yeah i see it too.
agent0smith
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I googled inscribed rectangles just to be sure.
Dido525
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Wait, why did you do a midpoint for the first rectangle?
Dido525
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Instead of a left one?
Dido525
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NOW this all makes sense :D .
agent0smith
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Inscribed rectangles can be either left or right endpoint
agent0smith
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or midpoint
agent0smith
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@jennychan12 you see how it's 26 now? the height of the first is f(1), next is f(2), then f(4)
Dido525
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I would give you my medal if I could :P .
agent0smith
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haha :) this whole confusion was all over the meaning of the word inscribed :D
Dido525
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Agreed! :D . But hey, this might be n my final so it was good for me too :) .
agent0smith
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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
Dido525
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She is offline :/ .
Dido525
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@jada1701877 : You should have given that to agent.
jennychan12
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thanks so much! and yeah, i went off at like 11 cuz i went to sleep. :)