Simpsons Rule: Calc. area of earth to be excavated, given int[1,5](1/x^2)dx w/ 8 inte

[JimCrown]

Using Simpson's area rule, calculate the area of earth to be excavated given the integral:

b

f(x)dx

a

Where a=1, b=5, and f(x)=1/x^2

do this using 8 intervals

5-1/8 = 0.5

x 2/√x

1 2
1.5 1.633
2 1.414
2.5 1.265
3 1.155
3.5 1.069
4 1
4.5 0.943
5 0.894


1.633 + 1.265 + 1.069 + 0.943 x 4 = 19.64
2 + 0.894 =2.894
1.414+ 1.155 + 1 =3.569 x2 =7.138

1/3 x 0.5 [2.894 + 9.64 + 7.138] = 8.701 to 3dp

Would this be the right working out/method and answer.

 

[stapel]

Using Simpson's area rule, calculate the area of earth to be excavated given the integral:

int [from a to b] f(x) dx

where a=1, b=5, and f(x)=1/x^2

Do this using 8 intervals.

My working:

5-1/8 = 0.5

I think you mean "(5 - 1)/8":

. . . . .\(\displaystyle \dfrac{5\, -\, 1}{8}\, =\, \dfrac{4}{8}\, =\, \dfrac{1}{2}\)

...or 0.5. (Grouping symbols make a big difference!)

x 2/√x

1 2
1.5 1.633
2 1.414
2.5 1.265
3 1.155
3.5 1.069
4 1
4.5 0.943
5 0.894

I think the above is meant to be a table...? But I'm not sure what is meant by "2/√x"...? Where did this come from?

Using the standard formula for Simpson's Rule (here), we have:

. . . . .x-values: 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0

. . . . .function: f(x) = 1/x^2

. . . . .y-values: f(1.0), f(1.5), f(2.0), f(2.5), f(3.0), f(3.5), f(4.0), f(4.5), f(5.0)

Of course, all values should be held "exact" until the very end, so decimals would come into play, if at all, only after the summation, which would be:

. . . . .\(\displaystyle \left(\dfrac{1/2}{3}\right)\, \left(\dfrac{1}{(1.0)^2}\, +\, 4\, \cdot\, \dfrac{1}{(1.5)^2}\, +\, 2\, \cdot\, \dfrac{1}{(2.0)^2}\, +\, 4\, \cdot\, \dfrac{1}{(2.5)^2}\, +\, 2\, \cdot\, \dfrac{1}{(3.0)^2}\, +\, 4\, \cdot\, \dfrac{1}{(3.5)^2} +\, 2\, \cdot\, \dfrac{1}{(4.0)^2}\, +\, 4\, \cdot\, \dfrac{1}{(4.5)^2}\, +\, \dfrac{1}{(5.0)^2}\right)\)

But this does not gibe with what you've posted...? Kindly please reply with clarification of (what I think are) your y-values. Thank you!

 

[JimCrown]

I think you mean "(5 - 1)/8":

. . . . .\(\displaystyle \dfrac{5\, -\, 1}{8}\, =\, \dfrac{4}{8}\, =\, \dfrac{1}{2}\)

...or 0.5. (Grouping symbols make a big difference!)


I think the above is meant to be a table...? But I'm not sure what is meant by "2/√x"...? Where did this come from?

Using the standard formula for Simpson's Rule (here), we have:

. . . . .x-values: 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0

. . . . .function: f(x) = 1/x^2

. . . . .y-values: f(1.0), f(1.5), f(2.0), f(2.5), f(3.0), f(3.5), f(4.0), f(4.5), f(5.0)

Of course, all values should be held "exact" until the very end, so decimals would come into play, if at all, only after the summation, which would be:

. . . . .\(\displaystyle \left(\dfrac{1/2}{3}\right)\, \left(\dfrac{1}{(1.0)^2}\, +\, 4\, \cdot\, \dfrac{1}{(1.5)^2}\, +\, 2\, \cdot\, \dfrac{1}{(2.0)^2}\, +\, 4\, \cdot\, \dfrac{1}{(2.5)^2}\, +\, 2\, \cdot\, \dfrac{1}{(3.0)^2}\, +\, 4\, \cdot\, \dfrac{1}{(3.5)^2} +\, 2\, \cdot\, \dfrac{1}{(4.0)^2}\, +\, 4\, \cdot\, \dfrac{1}{(4.5)^2}\, +\, \dfrac{1}{(5.0)^2}\right)\)

But this does not gibe with what you've posted...? Kindly please reply with clarification of (what I think are) your y-values. Thank you!

My teacher taught me to do that table. The y values are:

2
1.633
1.414
1.265
1.155
1.069
1
0.943
0.894

I was told do The n intervals in the formula 2 divided the squareroot of x term. So,2 divided by square root of 1 is 1

 

[stapel]

I was told do The n intervals in the formula 2 divided the square root of x term.

Why? On what logical basis are you changing the function?

When you reply, please include a clear statement of your class' version of Simpson's Rule, as this might help us figure out what's going on. Thank you!

 

[JimCrown]

Why? On what logical basis are you changing the function?

When you reply, please include a clear statement of your class' version of Simpson's Rule, as this might help us figure out what's going on. Thank you!

I do not know why. I was just taught this.

b

f(x) dx 1/3 (d) [(1st ordinate + last ordinate) + 4 (sum of even ordinates) + 2 (sum of remaining odd ordinates)]

a


we assume a parabolic curve

 

[stapel]

I do not know why. I was just taught this.

b

f(x) dx 1/3 (d) [(1st ordinate + last ordinate) + 4 (sum of even ordinates) + 2 (sum of remaining odd ordinates)]

a


we assume a parabolic curve

But what you've posted here used f(x), not the fourth root of f(x), as you've used in your previous posts. What they gave you (above) matches what is at the link you got earlier and what is in an earlier reply. These all agree, so try using this instead of the square-root whatever that somebody told you.

 

[JimCrown]

1 1.5 2 2.5 3 3.5 4 4.5 5	1 0.444 0.25 0.16 0.111 0.082 0.063 0.049 0.04

1 + 0.04 = 1.04
0.444 + 0.16 + 0.082 + 0.49 x 4 = 2.94
0.25 + 0.111 + 0.063 x 2 = 0.848

1/3 x 0.5 [1.04 + 2.94 + 0.848] = 0.805 to 3dp

But what you've posted here used f(x), not the fourth root of f(x), as you've used in your previous posts. What they gave you (above) matches what is at the link you got earlier and what is in an earlier reply. These all agree, so try using this instead of the square-root whatever that somebody told you.

 

[stapel]

x	f(x) = 1/x2
1 1.5 2 2.5 3 3.5 4 4.5 5	1 0.444 0.25 0.16 0.111 0.082 0.063 0.049 0.04

Yes and no. Yes, you're now using the correct function, being the "f(x) = y = 1/x²" that they'd given you. But, no, by truncating a bunch of decimals at two or three places, one thing you can not say is that your answer is "accurate to three decimal places".

Instead, to get the answer expected, use the exact form of each of the values. For instance:

. . .\(\displaystyle \mbox{For }\, x\, =\, 1.5\, =\, \frac{3}{2}:\)

. . . . .\(\displaystyle y\, =\, f\left(\frac{3}{2}\right)\, =\, \dfrac{1}{\left(\dfrac{3}{2}\right)^2}\, =\, \dfrac{1}{\left(\dfrac{9}{4}\right)}\, =\, \dfrac{4}{9}\)

When you add up all of the exact values, you'll then get the exact value of the integral, according to Simpson's Rule. At that point, if you want to, you can find the decimal approximation and then round to whatever number of decimal places you prefer.

Note: If your approximate answer is meant to explain that the instructions told you to round the final answer to three decimal places, then you'll definitely need to do the "exact" work explained and demonstrated above.