Sums-Integrals problem 02.mws

1. Exploring area.

Consider the region that lies under the curve y = 4*x-x^3 from x = 0 to x = 2 . Maple is a powerful graphing utility and we can get it to sketch this region by writing:

> plot(4*x-x^3,x=0..2,filled=true);

[Maple Plot]

We may estimate that the area of this region is approximately 3 square units, since it is roughly the shape of a triangle with base 2 and height 3. Let's see how we might use mathematics to find the exact area of this region. Since ancient times people have been using polygonal shapes (such as triangles) to approximate areas such as these. A very tractable method involves using rectangles to approximate areas. To get Maple to help us visualize the rectangles we will use 10 rectangles, whose height is determined by using the right-hand endpoint of each of the 10 intervals, hence the command rightbox below.

> with(student):

> rightbox(4*x-x^3,x=0..2,10);

[Maple Plot]

A reasonable approximation to the area of the region can be obtained by adding the areas of these 10 rectangles. Notice each of the rectangles has a base of 0.2 and a height which is given by the expression 4*x-x^3 evaluated at each of the 10 right-hand endpoints. So if we let f(x) = 4*x-x^3 then we could write for our approximation .2*f(.2)+.2*f(.4) + ... + .2*f(2) . We would like to point out two things here. First, this sum follows a pattern which we can exploit enabling us to rewrite the sum using sigma notation.

> Sum(0.2*f(0.2*i),i=1..10);

Sum(.2*sqrt(1-sqrt(2-sqrt(3-.2*i))),i = 1 .. 10)

At this point, we haven't let Maple know what f is, so when we ask it to find the sum (by using a little s)  

> sum(0.2*f(0.2*i),i=1..10);

.8400237189

we are back to what we had before translating to sigma notation. We know define f for Maple.

> f := x -> 4*x-x^3;

f := proc (x) options operator, arrow; 4*x-x^3 end ...

And now we can obtain the sum of the 10 rectangles.

> Sum(0.2*f(0.2*i),i=1..10);

Sum(.16*i-.16e-2*i^3,i = 1 .. 10)

> sum(0.2*f(0.2*i),i=1..10);

3.960000000

Secondly, there is a quicker way with Maple to get this sum by writing.

> rightsum(4*x-x^3,x=0..2,10);

1/5*Sum(4/5*i-1/125*i^3,i = 1 .. 10)

> evalf(rightsum(4*x-x^3,x=0..2,10));

3.960000000

While this second way is nice, don't ignore the first since it helps you to develop a habit of mind that will be very useful in working homework problems and understanding the ideas behind this course.

Submission:

(a) Just as we have done above for the case of n = 10 rectangles, do the same with n = 20 and n = 30 rectangles. That is, use Maple's rightbox command to construct the rectangles that will approximate the area of the region, find the sum of the areas of the rectangles by observing a pattern and using the Sum command, and finally find the sum by using the rightsum command (which should serve as a check of your work with the Sum command).

(b) Use the Maple commands, leftbox and leftsum with n=30 on the above function.

(c) Repeat with middlebox and middlesum.

(d) Which do you think is most accurate, rightsum, middlesum or leftsum?

Submission worksheet: