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Help! P3 Integration Q (Maths)
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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
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
(Original post by confusedguy234)
3/4e^2 I get
3/4e^2 I get
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#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
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
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
(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)
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)
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#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
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
(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).
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