Table: Time (weeks) 0 2 6 10
Level 210 200 190 180
The table above gives the level of a person’s cholesterol at different times during a 10-week treatment period. What is the average level over this 10-week period obtained by using a trapezoidal approximation with the subintervals [0, 2], [2, 6], and [6, 10] ?
the area would, of course, be the sum of the areas of the trapezoids, which you surely know how to calculate:
A = (210+200)/2 * 2 + (200+190)/2 * 4 + (190+180)/2 * 4
and the average value is A/10
Well, the cholesterol level seems to be dropping over time. Maybe they had a falling out with their favorite cheeseburger joint. Anyway, to find the average level over the 10-week period using the trapezoidal approximation, we'll need to calculate the area under the curve.
Let's divide the time period into three subintervals: [0, 2], [2, 6], and [6, 10]. The cholesterol levels at these time points are 210, 200, 190, and 180 respectively. Since we have decreasing levels, it's like going down a slide... or in this case, a cholesterol slide.
Using the trapezoidal formula, we can calculate the area under each subinterval and then sum them up. For the first subinterval, the average cholesterol level is (210 + 200) / 2 = 205. For the second subinterval, the average is (200 + 190) / 2 = 195. Finally, for the last subinterval, the average is (190 + 180) / 2 = 185.
Now, let's calculate the area of each trapezoid. The area of each trapezoid is the average value multiplied by the width of the interval. So, for the first trapezoid, the area is 205 * 2 = 410. For the second trapezoid, the area is 195 * 4 = 780. And for the last trapezoid, the area is 185 * 4 = 740.
Adding up the areas of these three trapezoids, we get 410 + 780 + 740 = 1930. Finally, we divide this sum by the total width of the interval, which is 10 - 0 = 10. So, the average level over the 10-week period obtained by using the trapezoidal approximation is 1930 / 10 = 193.
In conclusion, the average cholesterol level over the 10-week period is approximately 193, which means our subject is moving in the right direction. Keep up the good work and lay off the cheeseburgers!
To find the average level over the 10-week period using a trapezoidal approximation, we need to calculate the area under the curve formed by the data points in the table.
First, we will calculate the area of each trapezoid using the formula for the area of a trapezoid:
Area = (base1 + base2) * height / 2
For the first trapezoid, the bases are the levels at time 0 and time 2, and the height is 2 (the width of the subinterval [0, 2]).
Area1 = (210 + 200) * 2 / 2 = 410
For the second trapezoid, the bases are the levels at time 2 and time 6, and the height is 4 (the width of the subinterval [2, 6]).
Area2 = (200 + 190) * 4 / 2 = 780
For the third trapezoid, the bases are the levels at time 6 and time 10, and the height is 4 (the width of the subinterval [6, 10]).
Area3 = (190 + 180) * 4 / 2 = 740
Next, we will add up the areas of all three trapezoids and divide by the total width (10 weeks) to find the average level.
Average level = (Area1 + Area2 + Area3) / total width
= (410 + 780 + 740) / 10
= 1930 / 10
= 193
Therefore, the average level over the 10-week period obtained by using a trapezoidal approximation with the subintervals [0, 2], [2, 6], and [6, 10] is 193.
To find the average level over the 10-week period using a trapezoidal approximation, follow these steps:
1. Calculate the area of each trapezoid formed by the data in the table.
- For the first trapezoid ([0, 2]), use the formula: (b1 + b2) * h / 2, where b1 and b2 are the cholesterol levels at time 0 and time 2, and h is the width of the interval (2 - 0 = 2).
- (210 + 200) * 2 / 2 = 410
- For the second trapezoid ([2, 6]), use the same formula: (b1 + b2) * h / 2, where b1 and b2 are the cholesterol levels at time 2 and time 6, and h is the width of the interval (6 - 2 = 4).
- (200 + 190) * 4 / 2 = 780
- For the third trapezoid ([6, 10]), use the same formula: (b1 + b2) * h / 2, where b1 and b2 are the cholesterol levels at time 6 and time 10, and h is the width of the interval (10 - 6 = 4).
- (190 + 180) * 4 / 2 = 700
2. Add up the areas of all the trapezoids.
- 410 + 780 + 700 = 1890
3. Divide the total sum by the number of subintervals to find the average level.
- 1890 / 3 = 630
Therefore, the average level over the 10-week period, obtained by using a trapezoidal approximation, is 630.