Calculating distance from speed

[GaussianSurface]

Homework Statement

The speed of a runner increased during the first three seconds of a race. Her speed at half-second intervals is given in the table. Find lower and upper estimates for the distance that she traveled during these three seconds.
It follows the image's square.

Homework Equations

I found that:
Σf(x1) (Δx) + f(x2) (Δx) ... f(xn) (Δx)
and by relating it to the distance which is the d = velocity ⋅ time, I can assume that it is similar

The Attempt at a Solution

I found the lower and a friend of mine the upper which is correct but I still have doubts about how to compute the upper distance.
For the lower, since I know that it changes 1/2 in 3 seconds I can replace it by a Δt=0.5 and multiply the velocities by Δt which is the change it suffers constantly.
So (0)(0.5) + (6.3)(0.5) + (9.5)(0.5) + (13.3)(0.5) + (17)(0.5) + (18.8)(0.5) = 32.45 ft which is the LOWER and it is actually correct, I got all the points but when I am doing the ''UPPER'' I start multiplying all the velocities by all the times which I think I'm not correct and here is my first question, when I multiply all the times by all the velocities, am I getting the full distance in 3 seconds? I know it is not on the task but I just wanted to know.
(0)(0.5) + (6.3)(0.5) + (9.5)(0.5) + (13.3)(0.5) + (17)(0.5) + (18.8)(0.5) + 20.9(3.0) = 42.9 ft
And the most important question why for the ''UPPER'' distance the last velocity is multiplied by Δt=0.5 and I get the UPPER, why just a velocity differentiates the LOWER from the UPPER, I will really appreciate you can answer my doubts.

 

[Mark44]

Homework Statement

The speed of a runner increased during the first three seconds of a race. Her speed at half-second intervals is given in the table. Find lower and upper estimates for the distance that she traveled during these three seconds.
It follows the image's square.

Homework Equations

I found that:
Σf(x1) (Δx) + f(x2) (Δx) ... f(xn) (Δx)
and by relating it to the distance which is the d = velocity ⋅ time, I can assume that it is similar

The Attempt at a Solution

I found the lower and a friend of mine the upper which is correct but I still have doubts about how to compute the upper distance.
For the lower, since I know that it changes 1/2 in 3 seconds I can replace it by a Δt=0.5 and multiply the velocities by Δt which is the change it suffers constantly.
So (0)(0.5) + (6.3)(0.5) + (9.5)(0.5) + (13.3)(0.5) + (17)(0.5) + (18.8)(0.5) = 32.45 ft which is the LOWER and it is actually correct, I got all the points but when I am doing the ''UPPER'' I start multiplying all the velocities by all the times which I think I'm not correct and here is my first question, when I multiply all the times by all the velocities, am I getting the full distance in 3 seconds? I know it is not on the task but I just wanted to know.
(0)(0.5) + (6.3)(0.5) + (9.5)(0.5) + (13.3)(0.5) + (17)(0.5) + (18.8)(0.5) + 20.9(3.0) = 42.9 ft
And the most important question why for the ''UPPER'' distance the last velocity is multiplied by Δt=0.5 and I get the UPPER, why just a velocity differentiates the LOWER from the UPPER, I will really appreciate you can answer my doubts.
View attachment 218314

Did you graph the points in the table?
The distance will be the area under the graph of velocity vs. time. The lower and upper estimates are probably obtained using rectangles that respectively underestimate and overestimate the are (i.e., distance).
Since the velocity appears to be steadily increasing, rectangles using the left endpoint of each time interval will underestimate the distance, and rectangles that use the right endpoint of each time interval will overestimate the distance. There are probably examples in your textbook that illustrate what I'm talking about.