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### Riemann Sums as Estimates of Definite Integral

```Date: 10/11/2003 at 16:45:36
From: Amanda
Subject: calculus

The velocity of a car is given by

v(t) = t^2 + 2t

where v(t)is measured in ft/sec and t is measured in seconds.

Estimate the distance traveled from t=1 to t=4 using 3 subdivisions.
How accurate is this estimate?

I have no idea what to do. Do I just plug in my answers to check the
estimate?
```

```
Date: 10/12/2003 at 09:53:44
From: Doctor Luis
Subject: Re: calculus

Hi Amanda,

To get the actual distance, you would integrate v(t) over the interval
from t=1 to t=4.

To estimate the distance, you would create rectangles based on values
of v(t) at the endpoints of the subdivisions, and add up the
individual areas of these rectangles.  This is called a "Riemann sum."

E---F
|   |
|   |      Each rectangle is the product
C---D   |      of a rate and a time, which gives
|   |   |      a distance.
A---B   |   |
rate  |   |   |   |
|   |   |   |
|   |   |   |
+------------
time

Note that if the curve being estimated passes through points A, C, and
E, the estimate will be too low ("lower Riemann sum").  If it passes
through points B, D, and F, the estimate will be too high ("higher
Riemann sum").

Since they want three subdivisions, the times go from t_0=1 to t_1=2,
t_1=2 to t_2=3, t_2=3 to t_3=4.  That is, the time interval (1,4) was
split into the three intervals (1,2), (2,3), (3,4).

The lower Riemann sum will be:

distance = sum_k  v(t_k)*(t_{k+1} - t_k)

In your case, it's easy since there are only three subdivisions:

distance = v(1)*(2-1) + v(2)*(3-2) + v(3)*(4-3)

The answer is in feet since v(t) is in ft/sec and t is in sec.  This
underestimates the true distance.  To estimate the error, you have to
calculate the upper Riemann sum, which overestimates the true distance.

The upper Riemann sum will be:

distance = sum_k  v(t_{k+1})*(t_{k+1} - t_k)

= v(2)*(2-1) + v(3)*(3-2) + v(4)*(4-3)

The difference between these two will give you an estimate of the
error in your approximation:

e = |d_upper - d_lower|

= |   [v(2)*(2-1) + v(3)*(3-2) + v(4)*(4-3)]
- [v(1)*(2-1) + v(2)*(3-2) + v(3)*(4-3)]  |

Your accuracy is likely to be e/2 since you are using two estimates
that underestimate and overestimate the true distance.

To get a better estimate, you could evaluate a third Riemann sum
that's closer to the true distance.  You could probably get this by
evaluating v(t) not at the end points (t_k,t_{k+1}) of the k-th
subdivision, but at the midpoint m_k = (t_k+t_{k+1})/2:

d_midpoint = v(1.5)*(2-1) + v(2.5)*(3-2) + v(3.5)*(3-2)

As I said earlier, the actual distance is obtained with the integral

/ 4
|                             |4
d = | (t^2 + 2t)dt = (t^3/3 + t^2)|  = (4^3/3+4^2)-(1/3+1) = 36 ft
|                             |1
/ 1

You can see how much closer d_midpoint was to the correct answer than
either of d_lower or d_upper.

In case you haven't covered integrals yet, they are just Riemann sums
with really large numbers (millions, billions, trillions) of
subdivisions.  Technically, an integral is the limit of a Riemann sum
as the number of subdivisions approaches infinity.

I hope this helped!  Let us know if you have any more questions.

- Doctor Luis, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus

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