## jennychan12 3 years ago Integral Question. See below,

1. anonymous

O_o .

2. 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

3. jennychan12

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

4. jennychan12

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

5. anonymous

Wait.

6. anonymous

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

7. jennychan12

the problem doesnt say. i'm assuming left

8. jennychan12

cuz it says inscribed

9. anonymous

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

10. anonymous

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

11. jennychan12

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

12. anonymous

$\Delta(x) = 2$

13. anonymous

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

14. anonymous

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

15. jennychan12

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

16. anonymous

No wait, that's right.

17. 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) .

18. jennychan12

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

19. anonymous

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

20. jennychan12

26 or 28 ?

21. anonymous

It's 26.

22. anonymous

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

23. jennychan12

why? sorry can u explain?

24. jennychan12

oh.

25. jennychan12

so right hand it's 26 or..?

26. anonymous

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

27. anonymous

Well right hand is 76 actually.

28. anonymous

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

29. jennychan12

where'd the 26 come from? midpoint?

30. anonymous

No, midpoint is 46.

31. anonymous

But shall I explain left and right rectangles?

32. anonymous

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

33. anonymous

Shall I explain?

34. jennychan12

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

35. anonymous

26 isn't the correct answer.

36. 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...

37. jennychan12

oh.

38. anonymous

Wait hold on.

39. anonymous

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

40. anonymous

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

41. anonymous

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

42. jennychan12

yeah i did that... and got 28

43. anonymous

Which is absolutely correct.

44. anonymous

It CANNOT be 26 if you use left rectangles.

45. anonymous

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

46. jennychan12

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

47. anonymous

Yep, you teacher is wrong.

48. jennychan12

thanks for clarifying. :)

49. anonymous

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

50. jennychan12

yeah i know. i calculated it.

51. anonymous

Kk.

52. anonymous

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

53. anonymous

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

54. jennychan12

ohhhhhhh i think i got it they used average value

55. anonymous

LOL

56. 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)

57. anonymous

^ Reason why they they should specify the question.

58. jennychan12

my friend just assumed that inscribed meant left end points

59. 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.

60. anonymous

@jennychan12 : That's a flase assumption.

61. anonymous

false*

62. anonymous

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

63. 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.

64. 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 ^_^ .

65. anonymous

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

66. anonymous

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

67. anonymous

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

68. anonymous

and*

69. anonymous

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

70. anonymous

MUST*

71. 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.

72. anonymous

Conversation is too much for this PoSci major xD.

73. anonymous

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

74. anonymous

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

75. jennychan12

it's aiight.

76. agent0smith

Nobody drew a damn diagram :P

77. anonymous

Sigh...

78. anonymous

Diagram was drawn in the other thread.

79. jennychan12

^ haha yes

80. anonymous

|dw:1355899568667:dw|

81. anonymous

Something like that.

82. jennychan12
83. agent0smith

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

84. anonymous

LOL

85. anonymous

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

86. jennychan12

yeah i got really confused and tried to do another one to resolve my confusion... :D cuz that other person made me confused

87. anonymous

|dw:1355899646424:dw|

88. anonymous

|dw:1355899717852:dw|

89. anonymous

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

90. agent0smith

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

91. anonymous

Damn question.

92. anonymous

It's is 28 though without a doubt.

93. anonymous

More than likely your teacher made an error.

94. jennychan12

yeah

95. agent0smith

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

96. anonymous

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

97. agent0smith

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

98. jennychan12

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

99. jennychan12

@agent0smith can u explain???

100. anonymous

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

101. anonymous

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

102. anonymous

The previous thread came up with 26. this one says 28. let's meet half way and say it's 27...oh wait.

103. jennychan12

lol

104. 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

105. agent0smith

How'd you get 28, @Dido525

106. jennychan12

left endpoint is 28

107. anonymous

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

108. 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.

109. agent0smith

That is not an inscribed rectangle.

110. 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?

111. anonymous

But f(0) is inscribed.

112. agent0smith

I'll draw it jenny. One sec.

113. anonymous

I want to see this too.

114. 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.

115. anonymous

Isn't that the same thing as inscribed?

116. anonymous

Inscribed meaning within.

117. agent0smith

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

118. jennychan12

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

119. anonymous

|dw:1355900439887:dw|

120. anonymous

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

121. anonymous

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

122. agent0smith

123. anonymous

Ohh shoot! O_o .

124. agent0smith

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

125. anonymous

Ahh, NOW this makes sense.

126. agent0smith

hahah :)

127. anonymous

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

128. jennychan12

ok yeah i see it too.

129. agent0smith

I googled inscribed rectangles just to be sure.

130. anonymous

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

131. anonymous

Instead of a left one?

132. agent0smith

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

133. anonymous

NOW this all makes sense :D .

134. agent0smith

Inscribed rectangles can be either left or right endpoint

135. agent0smith

or midpoint

136. agent0smith

@jennychan12 you see how it's 26 now? the height of the first is f(1), next is f(2), then f(4)

137. anonymous

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

138. agent0smith

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

139. anonymous

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

140. 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

141. anonymous

She is offline :/ .

142. anonymous

@jada1701877 : You should have given that to agent.

143. jennychan12

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