Integral Question.
See below,

- jennychan12

Integral Question.
See below,

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- schrodinger

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- anonymous

O_o .

- 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

sorry typed something wrong. and that 1 is supposed to be an A

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## More answers

- jennychan12

so
i got the integrated area to be 48, so answers either a b or c

- anonymous

Wait.

- anonymous

Wait, what kind of rectangles? Left, right, or midpoint?

- jennychan12

the problem doesnt say. i'm assuming left

- jennychan12

cuz it says inscribed

- anonymous

Well you are right. It's either A, B or C :D . Now can you do it out? :) .

- anonymous

\[\Delta(x)=\frac{ 6-0 }{ 3 }\]

- jennychan12

i'm confused
my friend says to use riemann sum.
what's the formula for that?

- anonymous

\[\Delta(x) = 2\]

- anonymous

In this particular example, a Riemann sum is basically the sum of the individual rectangles.

- anonymous

When you learn about definite intergals you will have a good idea about what they actually mean.

- jennychan12

is it \[\frac{ b-a }{ n } \sum_{i = 1}^{n} f(x) \] ??

- anonymous

No wait, that's right.

- 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

that's what i did but i got 28
and the answer's 26

- anonymous

Then you have to use diffrent kinds of rectangles because your answer is correct.

- jennychan12

26 or 28 ?

- anonymous

It's 26.

- anonymous

But if you use left rectangles then it's 28.

- jennychan12

why? sorry can u explain?

- jennychan12

oh.

- jennychan12

so right hand it's 26 or..?

- anonymous

So this is a very stupid question because it dosen't specify the type of rectangles.

- anonymous

Well right hand is 76 actually.

- anonymous

But it dosen't specify what type of rectangle. Left, right or midpoint.

- jennychan12

where'd the 26 come from? midpoint?

- anonymous

No, midpoint is 46.

- anonymous

But shall I explain left and right rectangles?

- anonymous

If you use left, you get 28. If you use right you get 76.

- anonymous

Shall I explain?

- jennychan12

wait so where did the 26 come from.
please explain. i'm so lost. :(

- anonymous

26 isn't the correct answer.

- 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

oh.

- anonymous

Wait hold on.

- anonymous

Yeah, 28 should be correct. I disagree with the answer.

- anonymous

If you use 3 subintervals you CANNOT get 26 in anyway.

- anonymous

The formula would be (2)(f(0)+f(2)+f(4))

- jennychan12

yeah i did that...
and got 28

- anonymous

Which is absolutely correct.

- anonymous

It CANNOT be 26 if you use left rectangles.

- anonymous

if you use right rectangles it's (2)(f(2)+f(4)+f(6)) which is 76.

- jennychan12

yeah i got 28 and 76 but i chose 28.
so my teacher is wrong. hmph.

- anonymous

Yep, you teacher is wrong.

- jennychan12

thanks for clarifying. :)

- anonymous

In reality the exact value of that integral is 48 but you will learn that later.

- jennychan12

yeah i know. i calculated it.

- anonymous

Kk.

- anonymous

Good 'ol Riemann's screwing with everybody's head.

- 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

ohhhhhhh i think i got it
they used average value

- anonymous

LOL

- 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

^ Reason why they they should specify the question.

- jennychan12

my friend just assumed that inscribed meant left end points

- 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

@jennychan12 : That's a flase assumption.

- anonymous

false*

- anonymous

@oldrin.bataku : If you use midpoints, it's 46 not 26.

- 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

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

@Dido525 you're right, 2(5) = 10 so it's actually 2(23) = 46 for the approximate area.

- 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

That's why we have the trapozoidal rule ;) . Ans simpsons rule :P .

- anonymous

and*

- anonymous

But yes @oldrin.bataku is correct. If they MUST be inscribed then you MUSt use left rectangles.

- anonymous

MUST*

- 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

Conversation is too much for this PoSci major xD.

- 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

Sorry that we turned this page into a discussion @jennychan12 .

- jennychan12

it's aiight.

- agent0smith

Nobody drew a damn diagram :P

- anonymous

Sigh...

- anonymous

Diagram was drawn in the other thread.

- jennychan12

^ haha yes

- anonymous

|dw:1355899568667:dw|

- anonymous

Something like that.

- jennychan12

if u wanna see
http://openstudy.com/study#/updates/50d15940e4b069abbb70d3c0

- agent0smith

that's the same thread as this one...

- anonymous

LOL

- anonymous

|dw:1355899651047:dw|
That would be incorrect since it does not enclose the curve.

- 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

|dw:1355899646424:dw|

- anonymous

|dw:1355899717852:dw|

- anonymous

If you use midpoints it still dosen't enclose the curve fully.

- agent0smith

Inscribed rectangles *should* mean rectangles within the curve, but it sounds like that didn't work, so the damn question is broken.

- anonymous

Damn question.

- anonymous

It's is 28 though without a doubt.

- anonymous

More than likely your teacher made an error.

- jennychan12

yeah

- agent0smith

inscribed rectangles would give 26, but i feel like that was already covered in this mess somewhere

- anonymous

You haven't done "work" problems in your calc class yet have you?

- agent0smith

Wait, why isnt the answer 26? 3 inscribed rectangles = area of 26.

- jennychan12

what do u mean?
kinetic energy? i do that in physics....

- jennychan12

@agent0smith can u explain???

- anonymous

there are "work" problems you use calculus for. Pray that you do not have to do them xD

- anonymous

If you use 3 inscribed rectangles it's 28 not 26.

- 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

lol

- 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

How'd you get 28, @Dido525

- jennychan12

left endpoint is 28

- anonymous

=(2) * ( f(0) + f(2) + f(4) )
= (2) * ( 2 + 2 + 10 )
= 28

- 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

That is not an inscribed rectangle.

- 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

But f(0) is inscribed.

- agent0smith

I'll draw it jenny. One sec.

- anonymous

I want to see this too.

- 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

Isn't that the same thing as inscribed?

- anonymous

Inscribed meaning within.

- agent0smith

No, inscribed means it falls completely within the curve, no jutting out over it.

- jennychan12

okay. i understand that f(1) = 1 but....
oh wait. i see where you're going with this....

- anonymous

|dw:1355900439887:dw|

- anonymous

But rectangles will never cover a curve fully, it's impossible.

- anonymous

Like, it covers it fully but it's an overestimate.

- agent0smith

##### 1 Attachment

- anonymous

Ohh shoot! O_o .

- agent0smith

Those are inscribed rectangles, they underestimate the area greatly of course.

- anonymous

Ahh, NOW this makes sense.

- agent0smith

hahah :)

- anonymous

But if it's an underestimate, it's not inscribed is it?

- jennychan12

ok yeah i see it too.

- agent0smith

I googled inscribed rectangles just to be sure.

- anonymous

Wait, why did you do a midpoint for the first rectangle?

- anonymous

Instead of a left one?

- agent0smith

Because that's the only way to have it inscribed.
http://en.wikipedia.org/wiki/Inscribed_figure

- anonymous

NOW this all makes sense :D .

- agent0smith

Inscribed rectangles can be either left or right endpoint

- agent0smith

or midpoint

- 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

I would give you my medal if I could :P .

- agent0smith

haha :) this whole confusion was all over the meaning of the word inscribed :D

- anonymous

Agreed! :D . But hey, this might be n my final so it was good for me too :) .

- 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

She is offline :/ .

- anonymous

@jada1701877 : You should have given that to agent.

- jennychan12

thanks so much! and yeah, i went off at like 11 cuz i went to sleep. :)

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