I need serious help with this problem distance and area

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Homework Help Overview

The discussion revolves around approximating the area under a curve using rectangles, specifically through the use of left, right, and midpoint sample points over the interval from x=0 to x=12.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of areas using rectangles and the significance of sample points. There is confusion about how to correctly apply left, right, and midpoint sample points in their calculations.

Discussion Status

Some participants have provided calculations for the areas using different sample points, while others are questioning the correctness of these calculations and the interpretation of sample points. There is an ongoing exploration of the definitions and methods involved.

Contextual Notes

Participants are working with specific values derived from a graph, and there is uncertainty about how these values were obtained and whether they align with the problem's requirements regarding sample points.

afcwestwarrior
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a. use 6 rectangles under the given graph of f from x=0 to x=12
(i)= L6 (sample points are left endpoints)
ii= r6 ('' '' '' '' right end point)
iii = m6 ('' '' '' '' are ,midpoints)

here is what i did, i sketched it already, y=f(x)

L6= 2 * 8.8 + 2 * 8 +... 2 * 1 = 69.6
r6 = 2 * 9 + 2 * 8.8 +... 2 * 4 = 85.6
m6= 1 (9) + 3(8.8) + 5 (8) ... 11 *4 = 222.4

is this right, i think i did it wrong because of it talks about sample points
 
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this is confusing, the only thing that confuses me is the sample points, what does that mean,
 
does anyone understand what I'm saying
 
It sounds like one is using rectangles to approximate the area under a curve, as opposed to trapezoids. However, as the width of the rectangles approaches zero, i.e. gets very small, then the sum of the areas of the rectangles approaches the exact area of under the curve.

The sample point is point x, for which the height, h = f(x)

The left sample points would be (0,f(0)), (2,f(2)), . . . (10,f(10)).

The right sample points would be (2,f(2)), (4,f(4)), . . . (12,f(12)), with the right side of the rectangle touching the curve f(x).

The mid sample points would be taken at 1, 3, 5, . . . 11 and with corresponding heights f(1), f(3), . . . f(11).

In all cases, if the 6 rectangles between 0 and 12 have uniform width, then all have width 2, and the area of each rectange = 2*f(x) where x is the sample point.
 
so i did it right,
except M6 equals 85.6
2*f(1) + 2*f(3) ... 2 f(11)=85.6
 
afcwestwarrior said:
a. use 6 rectangles under the given graph of f from x=0 to x=12
(i)= L6 (sample points are left endpoints)
ii= r6 ('' '' '' '' right end point)
iii = m6 ('' '' '' '' are ,midpoints)

here is what i did, i sketched it already, y=f(x)

L6= 2 * 8.8 + 2 * 8 +... 2 * 1 = 69.6
r6 = 2 * 9 + 2 * 8.8 +... 2 * 4 = 85.6
m6= 1 (9) + 3(8.8) + 5 (8) ... 11 *4 = 222.4

is this right, i think i did it wrong because of it talks about sample points
Well, where did you get those numbers, "8.8", "8", etc.? Presumably evaluated f(x) (actually got the values from the graph) for some values of x. How did you choose those values of x?
The problem said, first, "sample points are left endpoints" so you are expected to use the left end of each interval as the "x" value in f(x).
Then "sample points are right endpoints" so you should use the right end of each interval as "x" in f(x) for that interval.
Finally "sample points are midpoints" so you should use the midpoint of each interval as "x" in f(x) for that interval.
 

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