This discussion is closed.
Sugaray
Badges: 0
#1
Report Thread starter 15 years ago
#1
Hey, please could someone have a quick go at this Q because I only get half of the answer

Integrate (ln x)/(x^3) between limits e and 0

Now the answer is supposedly 1/4 - 3/(4e^2)
but I only get the -3/(4e^2) bit. I can't get the 1/4, I keep getting all these ln 0 and /0 which don't work so can someone see if they get this answer? Thanksssssss
0
confusedguy234
Badges: 0
#2
Report 15 years ago
#2
I get 3/16e^4
0
confusedguy234
Badges: 0
#3
Report 15 years ago
#3
Wait ive done it wrong
0
confusedguy234
Badges: 0
#4
Report 15 years ago
#4
I get -1/3e^2(1-1/4e)
0
confusedguy234
Badges: 0
#5
Report 15 years ago
#5
Wait Ive done it wrong again.
0
confusedguy234
Badges: 0
#6
Report 15 years ago
#6
3/4e^2 I get
0
Sugaray
Badges: 0
#7
Report Thread starter 15 years ago
#7
(Original post by confusedguy234)
3/4e^2 I get
same as i get but apparently there's still a 1/4 missing?? did u get ln 0 's and things/0 as well?
0
confusedguy234
Badges: 0
#8
Report 15 years ago
#8
Yeh, just did it yet again and I got -1/4e^2. This is really confusing me now.
0
confusedguy234
Badges: 0
#9
Report 15 years ago
#9
-3/4e^2. There we go..
0
theone
Badges: 0
Rep:
?
#10
Report 15 years ago
#10
(Original post by Sugaray)
Hey, please could someone have a quick go at this Q because I only get half of the answer

Integrate (ln x)/(x^3) between limits e and 0

Now the answer is supposedly 1/4 - 3/(4e^2)
but I only get the -3/(4e^2) bit. I can't get the 1/4, I keep getting all these ln 0 and /0 which don't work so can someone see if they get this answer? Thanksssssss
I think you mean e and 1.

Parts gives:

[-1/2 (ln(x))/x^2)] + 1/2 integral of 1/x^3. = -1/2e^2 + 1/4 - 1/4e^2 = 1/4 - 3/4e^2 = 1/4(1-3/e^2)
0
Sugaray
Badges: 0
#11
Report Thread starter 15 years ago
#11
(Original post by theone)
I think you mean e and 1.

Parts gives:

[-1/2 (ln(x))/x^2)] + 1/2 integral of 1/x^3. = -1/2e^2 + 1/4 - 1/4e^2 = 1/4 - 3/4e^2 = 1/4(1-3/e^2)
Well the book says 0 so I assume it's wrong. They've already made 2 printing errors on the same page stupid OCR >.< lol thanks
0
Jonny W
Badges: 8
Rep:
?
#12
Report 15 years ago
#12
(Original post by Sugaray)
Well the book says 0 so I assume it's wrong. They've already made 2 printing errors on the same page stupid OCR >.< lol thanks
For any k>0, the integral from k to e is

(e^2 - 3k^2 + 2e^2 ln(k))/(4e^2 k^2).

Call that I(k). Then I(1) = 1/4 - 3/(4e^2). Also, I(k) tends to -infinity as k tends to 0 from above. So the integral from 0 to e doesn't exist (the area under the curve is infinite).
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Bournemouth University
    Midwifery Open Day at Portsmouth Campus Undergraduate
    Wed, 18 Dec '19
  • The University of Law
    Open Day – GDL and LPC - Chester campus Postgraduate
    Sat, 4 Jan '20
  • University of East Anglia
    Mini Open Day Undergraduate
    Mon, 6 Jan '20

Did you vote in the 2019 general election?

Yes (278)
48.01%
No (66)
11.4%
I'm not old enough (235)
40.59%

Watched Threads

View All