Question - Use of trapezoidal rule (1 Viewer)

[Smile12345]

Hello All.

Could someone please show me the steps for the following question?

'Use one application of the trapezoidal rule to approximate:



Thanks in advance.

 

[rumbleroar]

1 (n) application = 2 (2n) sub intervals (two trapeziums) = 3 (2n+1) function values

1. draw up a table with "x" on the first row and "1/x" on the second row
2. use (b-a)/n to calculate the spacing in between each sub-interval, in this case, the spacing is equal to one. You want 3 function values, so x = 2, 3, 4.
3. sub into f(x) (1/x in this case) and fill out your table
4. use formula: A = h/2 (y0 + y(last) + 2(y(rest))
y0 = first y value that comes out when you sub the first limit in (in this case 2)
ylast = last y value that comes out when you sub the last limit in (in this case 4)
yrest = all the y values in between that aren't y0 or ylast (in this case 3)

note: h = (b-a)/n

 

[Smile12345]

1. draw up a table with "x" on the first row and "1/x" on the second row
2. use (b-a)/n to calculate the spacing in between each sub-interval, in this case, the spacing is equal to one. You want 3 function values, so x = 2, 3, 4.
3. sub into f(x) (1/x in this case) and fill out your table
4. use formula: A = h/2 (y0 + y(last) + 2(y(rest))
y0 = first y value that comes out when you sub the first limit in (in this case 2)
ylast = last y value that comes out when you sub the last limit in (in this case 4)
yrest = all the y values in between that aren't y0 or ylast (in this case 3)

note: h = (b-a)/n

Thanks for your help... I've tried that again but I'm not getting their answer of 3/4...

So the height = 1 as you said...

Then for
x = 2, y = 1/2,
x = 3, y = 1/3
x = 4, y = 1/4

So I've subbed it in as 2/2 (1/2 + 1/4 + 2(1/3) = 17/12

What am I doing wrong??

 

[QZP]

1 (n) application = 2 (2n) sub intervals (two trapeziums) = 3 (2n+1) function values

1. draw up a table with "x" on the first row and "1/x" on the second row
2. use (b-a)/n to calculate the spacing in between each sub-interval, in this case, the spacing is equal to one. You want 3 function values, so x = 2, 3, 4.
3. sub into f(x) (1/x in this case) and fill out your table
4. use formula: A = h/2 (y0 + y(last) + 2(y(rest))
y0 = first y value that comes out when you sub the first limit in (in this case 2)
ylast = last y value that comes out when you sub the last limit in (in this case 4)
yrest = all the y values in between that aren't y0 or ylast (in this case 3)

note: h = (b-a)/n

nope

 

[Smile12345]

Is anyone able to help me please? I'm unsure now!

 

[braintic]

It is only one application, so it should be easy.

Draw the diagram if you don't see it.

Draw y=1/x.
Draw the ordinates x=2 [ from (2,0) to (2, 1/2) ] and x=4 [ from (4,0) to (4, 1/4) ]
Join (2, 1/2) to (4, 1/4)

You should be able to find the area of the trapezium formed. (I hope you know how to find the area of a trapezium).



Once you've done it, you should be able to generalize :

Area (integral) = (h/2) [ f(a) + f(b) ], where h = b-a

where a and b in our example were 2 and 4, and h(x) was 1/x.



The longer formula given previously is a generalization when multiple trapezia are used, rather than adding up separate areas. It IS useful for those questions, but it is unnecessarily complicated for just one trapezium.

EDIT: Actually the first line in the previous formula was not correct. It referred to Simpson's rule.
You should be getting 2/2 (1/2 + 1/4)

 

[Smile12345]

It is only one application, so it should be easy.

Draw the diagram if you don't see it.

Draw y=1/x.
Draw the ordinates x=2 [ from (2,0) to (2, 1/2) ] and x=4 [ from (4,0) to (4, 1/4) ]
Join (2, 1/2) to (4, 1/4)

You should be able to find the area of the trapezium formed. (I hope you know how to find the area of a trapezium).



Once you've done it, you should be able to generalize :

Area (integral) = (h/2) [ f(a) + f(b) ], where h = b-a

where a and b in our example were 2 and 4, and h(x) was 1/x.



The longer formula given previously is a generalization when multiple trapezia are used, rather than adding up separate areas. It IS useful for those questions, but it is unnecessarily complicated for just one trapezium.

EDIT: Actually the first line in the previous formula was not correct. It referred to Simpson's rule.
You should be getting 2/2 (1/2 + 1/4)

Of course... Thanks heaps.