Illustrate the following and then answer the following :a. Compute the approximate area of the region bounded by the parabola and the x-axis. b. Describe the application of the Reiman sums?

Question

Procedure:

1. Draw a Cartesian plane on a ¼ size illustration board (use a scale of 1 centimeter.

2. Plot the graph of the parabola (x – 1)2 = – 16(y – 4) on the Cartesian plane. (Note that the

parabola is opening downward)

3. Divide the bounded region between the graph and the x-axis into approximately 8 equal parts.

4. Cut a rectangular strip (use bond paper or colored paper) that will exactly fit each part of the

bounded region.

5. Paste the strip on the graph from the left x-intercept all through out to the right x-intercept. Make

sure the middle top of the strip intersects the parabola.

6. After the entire region is covered by the strips, calculate the area of each rectangular strip and

write it on top.

7. You are now ready to perform the task below.

Reiman sums approximate area under a curve by accumulating the areas of rectangles. On a piece of

paper,

a. Compute the approximate area of the region bounded by the parabola and the x-axis.

b. Describe the application of the Reiman sums?

Answer 1

(x-1)^2 =-16(y-4)

y-4 = -(1/16)(x-1)^2

y = (1/16)(x-1)^2 +4 in vertex form with vertex = (1,4)

It's an downward opening parabola with maximum point = vertex = (1,4)

y = (-1/16)(x^2 -2x +1) +4

y intercept = (0, 3 15/16)

x intercepts are (-7,0) and (9,0)

calculate the Reiman sums on the right half of the parabloa, when you sum them up, multiply by two, as the parabola is symmetric about x=1

sum = 2{f(2) + f(4) + f(6) + f(8)] 2(3 15/16 + 3 7/16 + 2 5/16 + 15/16) = 2(10 5/8) = 21 7/8

multiply by for the other half area= total area under the parabola above x axis = 2(21 7/8) = 43 3/4

check this by taking the definite integral evaluated at -7 to 9

= 45., which is within 2 of the approximation by Reiman

another check on the answers is

calculate an upper bound of 4 times (9--7) = 4x16 = 64 = the area of a rectangle with base of 16 and height = 4. Then take half that which is less than the area under the parabola = 64/2 = 32

43 3/4 is between 32 and 64. half of (32+64) = 96/2 =48 = another approximation fairly close