Nt1310 Unit 6 Lab Report
Week 1 Lab
Name_________
Part I - Directions: For the function we are going to estimate the area under the curve using the trapezoidal rule where n = 5 by doing the following:
1. Divide the interval into 5 equal pieces. How long is each piece? This will represent the width of the rectangles we will use to estimate the area in gray above. What will the x-values be for the endpoint of each piece?
Width of rectangle = 0.4 .2
.8
1.6
2.4
3.2 2
2. Evaluate f(x) for x1 to x6 the endpoints. This will represent the bases of the trapezoids we will use to estimate the area under the curve.
(2*f (0.4)) (2*f(0.8)) (2*f(1.2)) (2*f(1.6)) (f(2))
3. Using the heights from part 2 and the width from part 1 draw the trapezoids onto the graph. Be sure that you measure the height from each endpoint. …show more content…
Based on the sketch will the estimate you get be an over or underestimate?
Why?
Based on the sketch my estimate will be an underestimate because the equation that I’m using is not very exact but it will be very close to the actual precise answer.
4. The Trapezoidal rule is . Plug the values you got in part II into this formula. (Hint: you should have 5 terms inside the parenthesis with the middle three terms all multipled by 2.) What estimate for the definite integral did you get?
For the definite integral I got 20/3 or 6.6667.
Part II: Simpson’s Rule
5. For the function we are going to estimate the area under the curve using the Simpson’s rule where n = 6 by doing the
following:
= .33333 1.666665
.66666
1.33334
2
2.66666
3.33334
2
(1/2) * (0.33333333333333)
2*f(0.33333333333333)
2*f(0.66666666666667)
2*f(1)
2*f(1.3333333333333)
2*f(1.6666666666667)
f(2)
The formula for Simpson’s Rule is . Plug the values above into Simpson’s Rule and calculate your approximation.
After plugin in the Simpson’s Rule I got 20/3 or 6.6667
6. Which formula, Trapezoidal or Simpson’s, gives the better approximation?
The Simpson’s rule gives better approximation.
Part III
1. Estimate the area under the curve in the interval [-3, 3] with n = 6 using the trapezoidal rule.
For the area I got 35. Δx = (-3 - 3)/6
Δx = -6/6
Δx = -1
2. Estimate the area under the curve in the interval [0, 4] with n = 8 using Simpson’s Rule.
Area = 10
Δx = (4 - 0)/8
Δx = 4/8
Δx = 0.5
3. Suppose an object with initial acceleration of 10 is dropped from 300 ft. above the earth, how far would it travel in the first 2 seconds of its journey? Assume initial velocity is zero. V = -64 ft/s
Name_________
Part I - Directions: For the function we are going to estimate the area under the curve using the trapezoidal rule where n = 5 by doing the following:
1. Divide the interval into 5 equal pieces. How long is each piece? This will represent the width of the rectangles we will use to estimate the area in gray above. What will the x-values be for the endpoint of each piece?
Width of rectangle = 0.4 .2
.8
1.6
2.4
3.2 2
2. Evaluate f(x) for x1 to x6 the endpoints. This will represent the bases of the trapezoids we will use to estimate the area under the curve.
(2*f (0.4)) (2*f(0.8)) (2*f(1.2)) (2*f(1.6)) (f(2))
3. Using the heights from part 2 and the width from part 1 draw the trapezoids onto the graph. Be sure that you measure the height from each endpoint. …show more content…
Based on the sketch will the estimate you get be an over or underestimate?
Why?
Based on the sketch my estimate will be an underestimate because the equation that I’m using is not very exact but it will be very close to the actual precise answer.
4. The Trapezoidal rule is . Plug the values you got in part II into this formula. (Hint: you should have 5 terms inside the parenthesis with the middle three terms all multipled by 2.) What estimate for the definite integral did you get?
For the definite integral I got 20/3 or 6.6667.
Part II: Simpson’s Rule
5. For the function we are going to estimate the area under the curve using the Simpson’s rule where n = 6 by doing the
following:
= .33333 1.666665
.66666
1.33334
2
2.66666
3.33334
2
(1/2) * (0.33333333333333)
2*f(0.33333333333333)
2*f(0.66666666666667)
2*f(1)
2*f(1.3333333333333)
2*f(1.6666666666667)
f(2)
The formula for Simpson’s Rule is . Plug the values above into Simpson’s Rule and calculate your approximation.
After plugin in the Simpson’s Rule I got 20/3 or 6.6667
6. Which formula, Trapezoidal or Simpson’s, gives the better approximation?
The Simpson’s rule gives better approximation.
Part III
1. Estimate the area under the curve in the interval [-3, 3] with n = 6 using the trapezoidal rule.
For the area I got 35. Δx = (-3 - 3)/6
Δx = -6/6
Δx = -1
2. Estimate the area under the curve in the interval [0, 4] with n = 8 using Simpson’s Rule.
Area = 10
Δx = (4 - 0)/8
Δx = 4/8
Δx = 0.5
3. Suppose an object with initial acceleration of 10 is dropped from 300 ft. above the earth, how far would it travel in the first 2 seconds of its journey? Assume initial velocity is zero. V = -64 ft/s