AB = ½ (0.2)(3.04 + 3.16)
AB = 0.62
Trapezium C:
b1 : g(x) = x2 + 3 b2 : g(x) = x2 + 3
g(0.4) = (0.4)2 + 3 g(0.6) = (0.6)2 + 3
= 3.16 = 3.36
AC = ½ (0.2)(3.16 + 3.36)
AC = 0.652
Trapezium D:
b1 : g(x) = x2 + 3 b2 : g(x) = x2 + 3
g(0.6) = (0.6)2 + 3 g(0.8) = (0.8)2 + 3
= 3.36 = 3.64
AD = ½ (0.2)(3.36 + 3.64)
AD = 0.7
Trapezium E:
b1 : g(x) = x2 + 3 b2 : g(x) = x2 + 3
g(0.8) = (0.8)2 + 3 g(1) = (1)2 + 3
= 3.64 = 4
AE = ½ (0.2)(3.64 + 4)
AE = 0.764
Total Area:
A = AA + AB + AC + AD + AE
A = 0.604 + 0.62 + 0.652 + 0.7 + 0.764
A = 3.34
With the help of technology, create diagrams showing and increasing number of trapeziums. For each diagram, find the approximation for the area.
What do you notice?
- First I calculated the area manually by hand. Then I tested these results with the use of technology, which in this case was the Riemann Sum application on a TI-84+ silver edition . Which I used to create the table above.
Observations:
The value of the sum continues to get more and more precise as the number of trapeziums under the curve increases.
Use the diagram below to find a general expression for the area under the curve of g, from x = 0 to x = 1, using n trapeziums.
∑ = ½ h [g(x0) + g(x1)] + ½ h [g(x1) + g(x2)] + ½ h [g(x2) + g(x3)] + ½ h [g(x3) + (g(xn)]
Two Trapeziums:
g(x) = x2 + 3
from x = 0 to x = 1
∑ = ½ h [g(x0) + g(x1)] + ½ h[g(x1) + g(xn)]
= ½ (0.5) (3 + 3.25) + ½ (0.5) (3.25 + 4)
= ½ (0.5) [3 + 4 + (2 ∗ 3.25)]
= ½ (0.5) [3 + 4 + 2(3.25)]
Use your results to develop the general statement that will estimate the area under any curve y = f(x) from x = a to x = b using n trapeziums. Show clearly how you developed your statement.
∑ = ½ h [g(x0) + g(x1)] + ½ h [g(x1) + g(x2)] + ½ h [g(x2) + g(x3)] + ½ h [g(x3) + (g(xn)]
h = ½ b – a = b – a = xn – x0
n 2n 2n
∑ = xn – x0 (g(x0) + g(xn) + 2(g(x1) + g(x2) + g(x3))
2n
= xn – x0 (g(x0) + g(xn) + 2[∑ g(xi)]
2n
Use your general statement, with eight trapeziums, to find approximations for these areas.
∑ = xn – x0 (g(x0) + g(xn) + 2[∑ g(xi)]
2n
= 3 – 1 [0.63 + 1.31 + 2 (0.73 + 0.83 + 0.91 + 1 + 1.08 + 1.16 + 1.24)]
2(8)
=1.98
∑ = xn – x0 (g(x0) + g(xn) + 2[∑ g(xi)]
2n
= 3 – 1 [2.85 + 4.5 + 2 (3.40 + 3.84 + 4.16 + 4.37 + 4.48 + 4.53 + 4.54)]
2(8)
= 8.24625
∑ = xn – x0 (g(x0) + g(xn) + 2[∑ g(xi)]
2n
= 3 – 1 [3 + 3 + 2(3.875 + 3.75 + 3 + 2 + 1.125 + 0.75 + 1.25)]
2(8)
= 4.6875
Find and compare these answers with your approximations. Comment on the accuracy of your approximations.
Calculator:
∫f(x)dx = 1.9806909
Calculator:
∫f(x)dx = 8.2597312
Calculator:
∫f(x)dx = 4.6666667
- My approximations are accurate because the answers that I originally got are very close to the calculator answers.
Use other functions to explore the scope and limitations of your general statement. Does it always work? Discuss how the shape of a graph influences your approximation.
- 4 Trapeziums – x = 0 to x = 1
y = x4 + 16x + 2
∑ = xn – x0 (g(x0) + g(xn) + 2[∑ g(xi)]
2n
x0 : g(x) = x4 + 16x + 2
g(0) = (0)4 + 16(0) + 2
= 2
xn : g(x) = x4 + 16x + 2
g(1) = (1)4 + 16(1)
= 19
x1 : g(x) = x4 + 16x + 2
g(0.25) = (0.25)4 + 16(0.25) + 2
= 6.00390625
x2: g(x) = x4 + 16x + 2
g(0.5) = (0.5)4 + 16(0.5) + 2
= 10.0625
x3: g(x) = x4 + 16x + 2
g(0.75) = (0.75)4 + 16(0.75) + 2
= 14.31640625
∑ = 1 – 0 (0 + 17 + 2[6.00390625 + 10.0625 + 14.31640625]
2(4)
= 9.861328125
Calculator:
∫f(x)dx = 10.2
y = x5 + 30
∑ = xn – x0 (g(x0) + g(xn) + 2[∑ g(xi)]
2n
x0 : g(x) = x5 + 30
g(0) = (0)5 + 30
= 30
xn : g(x) = x5 + 30
g(1) = (1)5 + 30
= 31
x1: g(x) = x5 + 30
g(0.25) = (0.25)5 + 30
= 30.000976563
x2: g(x) = x5 + 30
g(0.5) = (0.5)5 + 30
= 30.03125
x3: g(x) = x5 + 30
g(0.75) = (0.75)5 + 30
= 30.23730469
∑ = 1 – 0 (30 + 31 + 2[30.000976563 + 30.03125 + 30.23730469]
2(4)
= 30.19238281
Calculator:
Y= enter y = x5 + 25 2nd TRACE 7 ENTER graph displayed 0 ENTER
1 S ENTER S
∫f(x)dx = 30.166667
y = x + 2
4
∑ = xn – x0 (g(x0) + g(xn) + 2[∑ g(xi)]
2n
x0 : g(x) = x + 2
4
g(0) = (0) + 2
4
= ½
xn : g(x) = x + 2
4
g(1) = (1) + 2
4
= 3/4
x 1: g(x) = x + 2
4
g(0.25) = (0.25) + 2
4
= 0.5625
x2 : g(x) = x + 2
4
g(0.5) = (0.5) + 2
4
= 0.625
x3 : g(x) = x + 2
4
g(0.75) = (0.75) + 2
4
= 0.6875
∑ = 1 – 0 (1/2 + 3/4 + 2[0.5625 + 0.625 + 0.6875]
2(4)
= 0.625
Calculator:
∫f(x)dx = 0.625
Conclusion:
My original calculations from my general statement seemed to be very close to the answers that I later found using a calculator. I do not believe that the shapes of the graphs have an effect on the approximation