HSC question - e (1 Viewer)

[Saintly Devil]

(i) Evaluate the integral of 1/x from 1 to 2

(ii) Use Simpson's rule with 3 function values to approximate the above integral.

(iii) Use your results to parts (i) and (ii) to obtain an approximation for e. Give your answer correct to 3 decimal places.


I can do parts (i) and (ii), but I don't have a clue how to do (iii). Anyone?

 

[Lazarus]

Note that (i) gives you the integral in terms of e (a logarithm of base e, in fact).

And (ii) gives you the same integral as a number...



Equate!

 

[...]

integral of 1/x = ln x doesn't it???

 

[iambored]

Originally posted by ...
integral of 1/x = ln x doesn't it???

yep

 

[Huy]

Originally posted by iambored
yep

or Log(e)X, as Laz as stated above

 

[Fosweb]

Then just take the .6944 th root of 2 to get your approximation.

 

[iambored]

Originally posted by Huy
or Log(e)X, as Laz as stated above

same thing, u just wanted another post

 

[Huy]

Originally posted by iambored
same thing, u just wanted another post

I don't know what you're talking about!


...

 

[...]

Originally posted by Fosweb
Then just take the .6944 th root of 2 to get your approximation.

woah??

 

[iambored]

Originally posted by ...
woah??

i think he's talking crap, do what laz said

 

[Fosweb]

haha... sorry.
ok - let me explain.
i) You evaluate the first integral, which turns out to be ln(2)
ii) Simpsons rule gives you 25/36 = .6944

iii) Now you know these two answers are APPROXIMATELY equal, which is what the question is asking for. So equate them to find 'e' which is thus your only 'variable'.

So: ln(2) = 25/36 = 0.6944
Using log laws: 2 = e^0.6944
Therefore e = 0.6944root(2) = 2.713 to 3 dp.

Sorry i dont know the correct notation (ie how to write non-integer roots...i added an image instead.)

 

[iambored]

Originally posted by Fosweb
haha...
ok - let me explain.
i) You evaluate the first integral, which turns out to be ln(2)
ii) Simpsons rule gives you 25/36 = .6944

iii) Now you know these two answers are APPROXIMATELY equal, which is what the question is asking for. So equate them to find 'e' which is thus your only 'variable'.

So: ln(2) = 25/36 = 0.6944
Using log laws: 2 = e^0.6944
Therefore e = 0.6944root(2) = 2.713 to 3 dp.

Sorry i dont know the correct notation (ie how to qrite non-integer roots...)

ah well that makes sense, i thought u just chose a number, since i haven't done the question myself

 

[...]

i get it now (after the pic has been attached)...

 

[crazylilmonkee]

ohhhhhhh.......
didnt think of that

 

[Saintly Devil]

Originally posted by Lazarus
Note that (i) gives you the integral in terms of e (a logarithm of base e, in fact).

And (ii) gives you the same integral as a number...



Equate!

Ah, damn.
Thanks.