Subsections

• Rectangular Approximation
• Motion
• Exercises

Area Under a Curve and Antidifferentiation

In this lab you will study a link between finding the area under the graph of a function and antidifferentiating the function.

Rectangular Approximation

In the following simple approximation schemes, the area under the curve and above each subinterval is approximated by the area of a rectangle, with the height of the rectangle being chosen according to some rule. The rules we will be concerned with are as follows.

left endpoint rule

The height of the rectangle is the value of the function f(x) at the left-hand endpoint of the subinterval.

right endpoint rule

The height of the rectangle is the value of the function f(x) at the right-hand endpoint of the subinterval.

The Maple student package has commands for visualizing these two rectangular area approximations. To use them, you first must load the student package via the with command. Then try the commands given below. Make sure you understand the differences between the two different rectangular approximations.

  > with(student):

  > f:=x-> x^2 ;

  > rightbox(f(x),x=0..4);

  > leftbox(f(x),x=0..4);

There are also Maple commands leftsum and rightsum to sum the areas of the rectangles, see the examples below. Note the use of evalf to obtain numerical answers.

  > rightsum(f(x),x=0..4);

  > evalf(rightsum(f(x),x=0..4));

  > evalf(rightsum(x,x=0..2));

Unless specified, Maple will break the given interval into four subintervals. Below are some examples of how to change the number of subintervals used in the approximation.

  > evalf(rightsum(x,x=0..2,10));

  > evalf(rightsum(x,x=0..2,20));

  > evalf(rightsum(x,x=0..2,100));

Motion

You have seen in lecture that the position p(t) of an object can be recovered from the velocity v(t) if one knows position at some particular time. This is done by antidifferentiating v(t). However, there is another way to think about recovering the position function, based on finding ``area'' under the velocity function.

Suppose v(t)=t², p(0)=0, and you want to find p(1) (without knowing calculus). How could you do this? First, just try to find a good approximation of position at time 1. Notice that the only thing you really know about finding position from velocity is that if velocity is constant (V) and the object travels for time T, then the change in position is V T. This can be used to approximate p(1) since if the time interval [0,1] is divided into n subintervals, v(t) will be close to a constant in each subinterval (for large enough n), and so the change in position over a time interval of length $\frac{1}{n}$ is approximately $(t_i)^2 \frac{1}{n}$, where t_i is any particular time in our time interval. The total change in position can then be approximated by

$$\sum_{i=1}^n (i\frac{1}{n})^2 \frac{1}{n}$$

and in fact is equal to

$$\lim_{n \rightarrow \infty} \sum_{i=1}^n (i\frac{1}{n})^2 \frac{1}{n}.$$

Notice that this is just the area under the velocity function.

Exercises

1.

Use the left endpoint rule and then the right endpoint rule to approximate p(1) (n=10).

2.

Explain why velocity can be approximated by approximating the area under the acceleration function. Approximate v(10) if a(t)=10 and v(0)=50 (n=20) (note that the sum is more complicated now, since the interval is [0,10] instead of [0,1]. What needs to be changed?). Explain what would happen if acceleration were negative (would you still find area under the curve?).

3.

Approximate the position p(10) if p(0)=0, v(0)=50, and a(t)=-10 (n=20).