3) The table below gives selected values for the function f(x). Use a trapezoidal estimation, with 6 trapezoids to approximate the value of int f(x) dx from 1 to 2 . Give 3 decimal places for your answer.
x 1 1.1 1.2 1.5 1.7 1.9 2.0
f(x) 1 2 4 6 7 9 10
please help! I've been trying to use my notes and the usual rule but I can't seem to get it.
To approximate the value of ∫f(x) dx from 1 to 2 using trapezoidal estimation, follow these steps:
Step 1: Determine the width of each trapezoid by calculating Δx (change in x):
Δx = (b - a) / n
where a is the lower limit (1 in this case), b is the upper limit (2 in this case), and n is the number of trapezoids (6 in this case).
Δx = (2 - 1) / 6 = 1/6 = 0.1667
Step 2: Calculate the areas of all 6 trapezoids:
Trapezoid 1:
Base1 = f(1) = 1
Base2 = f(1.1) = 2
Area1 = Δx * (Base1 + Base2) / 2 = 0.1667 * (1 + 2) / 2 = 0.1667 * 3 / 2 = 0.2500
Trapezoid 2:
Base1 = f(1.1) = 2
Base2 = f(1.2) = 4
Area2 = Δx * (Base1 + Base2) / 2 = 0.1667 * (2 + 4) / 2 = 0.1667 * 6 / 2 = 0.5000
Trapezoid 3:
Base1 = f(1.2) = 4
Base2 = f(1.5) = 6
Area3 = Δx * (Base1 + Base2) / 2 = 0.1667 * (4 + 6) / 2 = 0.1667 * 10 / 2 = 0.8333
Trapezoid 4:
Base1 = f(1.5) = 6
Base2 = f(1.7) = 7
Area4 = Δx * (Base1 + Base2) / 2 = 0.1667 * (6 + 7) / 2 = 0.1667 * 13 / 2 = 1.0833
Trapezoid 5:
Base1 = f(1.7) = 7
Base2 = f(1.9) = 9
Area5 = Δx * (Base1 + Base2) / 2 = 0.1667 * (7 + 9) / 2 = 0.1667 * 16 / 2 = 1.3333
Trapezoid 6:
Base1 = f(1.9) = 9
Base2 = f(2.0) = 10
Area6 = Δx * (Base1 + Base2) / 2 = 0.1667 * (9 + 10) / 2 = 0.1667 * 19 / 2 = 1.5833
Step 3: Sum up all the areas to approximate the integral:
Approximation of ∫f(x) dx from 1 to 2 = Area1 + Area2 + Area3 + Area4 + Area5 + Area6
≈ 0.2500 + 0.5000 + 0.8333 + 1.0833 + 1.3333 + 1.5833
≈ 4.5833
Therefore, the approximate value of ∫f(x) dx from 1 to 2 using trapezoidal estimation with 6 trapezoids is 4.583 to 3 decimal places.
To approximate the value of a definite integral using the trapezoidal rule, you need to follow these steps:
1. Divide the interval [1, 2] into equal subintervals or trapezoids. In this case, you have 6 trapezoids since you are given 7 data points.
2. Calculate the width of each trapezoid by taking the difference between consecutive x-values:
- For the 1st trapezoid: h = 1.1 - 1 = 0.1
- For the 2nd trapezoid: h = 1.2 - 1.1 = 0.1
- For the 3rd trapezoid: h = 1.5 - 1.2 = 0.3
- For the 4th trapezoid: h = 1.7 - 1.5 = 0.2
- For the 5th trapezoid: h = 1.9 - 1.7 = 0.2
- For the 6th trapezoid: h = 2 - 1.9 = 0.1
3. Calculate the average height of each trapezoid by taking the sum of the corresponding f(x) values and dividing it by 2 (since each trapezoid has two heights):
- For the 1st trapezoid: (1 + 2) / 2 = 1.5
- For the 2nd trapezoid: (2 + 4) / 2 = 3
- For the 3rd trapezoid: (4 + 6) / 2 = 5
- For the 4th trapezoid: (6 + 7) / 2 = 6.5
- For the 5th trapezoid: (7 + 9) / 2 = 8
- For the 6th trapezoid: (9 + 10) / 2 = 9.5
4. Calculate the area of each trapezoid using the formula: Area = (b1 + b2) * h / 2, where b1 and b2 are the heights of the trapezoid and h is the width of the trapezoid.
- For the 1st trapezoid: Area = (1 + 1.5) * 0.1 / 2 = 0.15
- For the 2nd trapezoid: Area = (2 + 3) * 0.1 / 2 = 0.25
- For the 3rd trapezoid: Area = (4 + 5) * 0.3 / 2 = 3.15
- For the 4th trapezoid: Area = (6 + 6.5) * 0.2 / 2 = 1.3
- For the 5th trapezoid: Area = (7 + 8) * 0.2 / 2 = 1.5
- For the 6th trapezoid: Area = (9 + 9.5) * 0.1 / 2 = 0.95
5. Sum up the areas of all the trapezoids to get the approximate value of the integral:
- Approximate value = 0.15 + 0.25 + 3.15 + 1.3 + 1.5 + 0.95 = 7.30
Therefore, the approximate value of the integral int f(x) dx from 1 to 2 using the trapezoidal rule with 6 trapezoids is 7.300 (to 3 decimal places).
not sure what's the trouble. You have the base lengths and the heights. Too bad you didn't show your work. It's just
(1+2)/2 * 0.1 + (2+4)/2 * 0.1 + ...
for the whole list of six intervals