Follow the procedure and answer the following questions.

Question

Procedure:

•     Draw a Cartesian Plane ( Use a scale of 1 centimeter)

•     Plot the graph of the parabola (x-1)²= -16(y-4) on the Cartesian Plane

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

•      Draw a rectangle that will exactly fit each part of the  bounded region.

•      Put the rectangle from the left x-intercept all through out to the right x-intercept. Make sure the middle top of the rectangle intersects the parabola.

•      After the entire region is covered by the rectangle ,calculate the area of each rectangle and

write it on top.

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) is a parabola which opens downward

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

y = (-1/16)(x^2 -2x +1) + 64/16

y = -x^2/16 + x/8 - 1/16 + 64/16

y = -x^2/48+ x/8 + 63/16, y intercept = (0, 63/16) = (0, 3 15/16)

integral of y = -x^3/48 + x^2/16 + 63x/16

evaluate at the x intercepts where y=0

(x-1)^2 = -(16)(y-4) =(-16)(-4) = 64

x-1 = + or - sqr64 = + or - 8

x = 1+8 or 1-8 = 9 or -7 = x intercepts

-x^3/48 +x^2/16 + 63x/16 evaluated from -7 to 9

= -(9)^3/48 + (9)^2/16 +63(9)/16 -[-(-7)^3/48 + (-7)^2/16 + 63(-7)/16]

= -729/48 + 81/16 + 567/16 - 343/48 - 49/16 + 441/16

= -1072/48 + 1089/16 - 49/16

= -1072/48 +1089/16

=(-1072 + 3267)/48

= 2195/48 = about 45.73

with geometry of a rectangle and triangle:

4(16)/2 < A < 4(16)

32 < the parabola area above the x axis < 64

64+32 = 96

96/2 = 48 = about 45.73

Reiman sums should be in that general range

divide into 8 equal parts

since the parabola is symmetric about x=1, just calculate 1/2 the area, then multiply by 2

use 4 equal parts on the right side of the parabola

then the midpoints of each rectangle are 2,4,6 and 8 each with base =2

heights are f(2), f(4), f(6) and f(8) = 63/16, 55/16, 39/16 and 15/16

2(2)(63+55+39+15)/16 = 172/4= 43 cm^2 which is an underestimate by about 2.7