y = 1/2x^2-x + 3 for 0_<x_<6.
(b) Calculate the mid-ordinates of 5 strips between x = 1 and x = 6, and hence use the mid- ordinate rule to approximate the area under between x = 1, x = 6 and the x-axis.
(c) Assuming that the area determined by integration to be the actual area, calculate the percentage error in using the mid-ordinate rule.
To calculate the mid-ordinates of the 5 strips between x = 1 and x = 6, we need to find the values of y for each mid-point within that interval.
First, let's determine the width of each strip. The interval between x = 1 and x = 6 is 6 - 1 = 5. Since we want 5 strips, each strip will have a width of 5/5 = 1.
Next, let's find the mid-points within the interval. The mid-point of the first strip is (1 + 1)/2 = 1.5. The mid-point of the second strip is (1 + 2)/2 = 1.5. Using the same logic, we can determine that the mid-points for the remaining strips are 2.5, 3.5, 4.5, and 5.5.
Now, let's substitute these mid-points into the equation y = (1/2)x^2 - x + 3 to calculate the values of y for each mid-point:
y1 = (1/2)*(1.5)^2 - 1.5 + 3
y2 = (1/2)*(2.5)^2 - 2.5 + 3
y3 = (1/2)*(3.5)^2 - 3.5 + 3
y4 = (1/2)*(4.5)^2 - 4.5 + 3
y5 = (1/2)*(5.5)^2 - 5.5 + 3
Now, we can use the mid-ordinate rule to approximate the area under the curve between x = 1 and x = 6. The mid-ordinate rule formula states that the area under each strip is the width of the strip multiplied by the corresponding mid-ordinate value. Therefore, we can calculate the approximate area by adding up the areas of all the strips:
Area ≈ (1 * y1) + (1 * y2) + (1 * y3) + (1 * y4) + (1 * y5)
To calculate the percentage error in using the mid-ordinate rule, we need to compare it to the actual area determined by integration. You can calculate the actual area under the curve by integrating the function y = (1/2)x^2 - x + 3 between x = 1 and x = 6. Once you have the actual area, you can determine the percentage error using the following formula:
Percentage Error = (|Approximate Area - Actual Area| / Actual Area) * 100
Note that the exact value of the actual area will depend on the function being integrated, and you'll need to perform the integral to obtain it.