Question

Integral Question.
See below,

Answer 1

O_o .

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Wait.

Answer 6

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

Answer 7

the problem doesnt say. i'm assuming left

Answer 8

cuz it says inscribed

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

\[\Delta(x) = 2\]

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

No wait, that's right.

Answer 17

So yeah you get the idea. It's the sum of all the rectangles multiples by the change in x,(i.e the width) .

Answer 18

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

Answer 19

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

Answer 20

26 or 28 ?

Answer 21

It's 26.

Answer 22

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

Answer 23

why? sorry can u explain?

Answer 24

oh.

Answer 25

so right hand it's 26 or..?

Answer 26

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

Answer 27

Well right hand is 76 actually.

Answer 28

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

Answer 29

where'd the 26 come from? midpoint?

Answer 30

No, midpoint is 46.

Answer 31

But shall I explain left and right rectangles?

Answer 32

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

Answer 33

Shall I explain?

Answer 34

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

Answer 35

26 isn't the correct answer.

Answer 36

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...

Answer 37

oh.

Answer 38

Wait hold on.

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

yeah i did that...
and got 28

Answer 43

Which is absolutely correct.

Answer 44

It CANNOT be 26 if you use left rectangles.

Answer 45

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

Answer 46

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

Answer 47

Yep, you teacher is wrong.

Answer 48

thanks for clarifying. :)

Answer 49

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

Answer 50

yeah i know. i calculated it.

Answer 51

Kk.

Answer 52

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

Answer 53

I just hope I don't have to do it after my first year in uni :/ . AP calc helped so much though XD .

Answer 54

ohhhhhhh i think i got it
they used average value

Answer 55

LOL

Answer 56

$$A\simeq2(2^2-2(2)+2+4^2-2(4)+2+6^2-2(6)+2)\\\ \ \ =2(4-4+2+16-8+2+36-12+2)\\\ \ \ =2(38)=76$$ (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)

Answer 57

^ Reason why they they should specify the question.

Answer 58

my friend just assumed that inscribed meant left end points

Answer 59

@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.

Answer 60

@jennychan12 : That's a flase assumption.

Answer 61

false*

Answer 62

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

Answer 63

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.

Answer 64

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 ^_^ .

Answer 65

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

Answer 66

If it requires inscribed rectangles then the approximation must use left end-point... mid- and right- will not be entirely under the parabola.

Answer 67

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

Answer 68

and*

Answer 69

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

Answer 70

MUST*

Answer 71

@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.

Answer 72

Conversation is too much for this PoSci major xD.

Answer 73

I looked at Runge-kutta but it makes no sense XD . I am going much deeper into math next year so hopefully then :) .

Answer 74

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

Answer 75

it's aiight.

Answer 76

Nobody drew a damn diagram :P

Answer 77

Sigh...

Answer 78

Diagram was drawn in the other thread.

Answer 79

^ haha yes