Illustrate and answer the following.

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)

solve for y to find the vertex (h,k) where y=a(x-h)^2 + k

y = (-1/16)(x-1)^2 +4 is a downward opening parabola with vertex (1,4) = maximum point

(-1/16)(x-2x +1) + 64/16

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

= (-1/16)(x-2x-63)

=-(-1/16)(x-9)(x+7) zeros or x intercepts are -7 and 9 or the points (-7,0) and (9,0)

axis of symmetry is x= (9-7)/2 = x = 2/2 = 1 = the x coordinate of the vertex

plot those three points vertex and x intercepts, also the y intercept = 4-1/16 = 3 15/16 or the point (0, 3 15/16) or (0, 63/16) or (0, 3.9375)

since (0, 63/16) is on the parabola, another point on the parabola is symmetric, on the opposite side of x=1

the point (2, 63/16) or (2, 3.9375)

you have 5 points, plot them, and draw a smooth parabolic curve throught the points

put an arrow pointing down on the ends of the bottom of the parabola to indicate the parabola goes downward forever, without end, approaching negative infinity.

on the x axis, 9 and -7 are the x intercepts, divide that distance by 8 to get 8 intervals each with base = 2

-7 to -5, -5 to -3, -3 to -1, -1 to 1, 1 to 3, 3 to 5, 5 to 7, 7 to 9

save some time and just calculate the area under the parabola and above the x axis for just the right half, from x=1 to x=9, then multiply that area by 2, heights of each rectangle are f(2), f(4), f(6), f(8)

= 63/16, 55/16, 41/16, and 15/16

you should get close to the intergral of -x^2/16 + x/8 + 63/16 evaluated from 1 to 9, doubled