The region R enclosed between the graph of the function y=x^2−4x+5 and the x-axis for 0≤x≤4 is partitioned into four subintervals. Determine the areas of the four rectangles using MRAM and then approximate the area of R. Round areas to the nearest hundredth. Show and explain your work.
The answer I got for this problem is 9 square units.

9.00 square units

Step-by-step explanation:

The width of the interval is 4 − 0 = 4.  Divided by 4 equal subintervals, the width of each subinterval is 4/4 = 1.

The subintervals are:

0 ≤ x ≤ 1

1 ≤ x ≤ 2

2 ≤ x ≤ 3

3 ≤ x ≤ 4

MRAM is midpoint rectangular approximation method.  So we use the midpoints of each interval to find the height of the rectangle:

f(0.5) = (0.5)² − 4(0.5) + 5 = 3.25

f(1.5) = (1.5)² − 4(1.5) + 5 = 1.25

f(2.5) = (2.5)² − 4(2.5) + 5 = 1.25

f(3.5) = (3.5)² − 4(3.5) + 5 = 3.25

So the total approximate area is:

A = 3.25 + 1.25 + 1.25 + 3.25

A = 9.00

Graph: desmos.com/calculator/x8dcibqszo

9 units²

Step-by-step explanation:

f(x) = x² - 4x + 5

Four intervals, their midpoints and f(MP) for heights:

0≤x<1 --> 0.5 --> 3.25

1<x≤2 --> 1.5 --> 1.25

2<x≤3 --> 2.5 --> 1.25

3<x≤4 --> 3.5 --> 3.25

All four intervals have a width of 1 unit

Sum of areas = total area

= 1×3.25 + 1×1.25 + 1×1.25 + 1×3.25

= 9

answer:

it is general formula that is used to write value to its excet position

a11 a12 a13

a21 a22 a23

it is the example of order 2 by 3 matrix in this

a11 means value occur in 1 st row 1st column

a12 means value occurs in 1st row 2nd column

a13 means value occur in 1st row and 3rd column

a21 means value occurs in 2nd row and 1st column

a22 means value occurs in 2nd row and in 2nd column

a23 means value occurs in 2nd row and in 3rd column