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# integrating e watch

1. Ok, I have to evaluate this:
∫(x - 1/x )dx within the boundaries of 6 and 3 (I dont know how to make them float by the integral sign!)
Im not really sure what form I should be leaving my answer in...do I do it in decimals, or leave ln there, or what? Please could someone solve it for me so I can see what to do.
Thank you very much
franks xxx
2. (Original post by franks)
Ok, I have to evaluate this:
∫(x - 1/x )dx within the boundaries of 6 and 3 (I dont know how to make them float by the integral sign!)
Im not really sure what form I should be leaving my answer in...do I do it in decimals, or leave ln there, or what? Please could someone solve it for me so I can see what to do.
Thank you very much
franks xxx
3. ok, looks like i havent learnt about some of the rules about e and logs.
Hmm, third line, how do you get -ln6 + ln3 = ln 0.5
how does ln0.5 = -ln 2
Sorry, the answers to these are probably really obvious, but ive only had half a lesson on the whole of the 'e' stuff and I havent bought myself a text book yet!
Thank you!
4. lnA - lnB = ln(A/B)
lnA+lnB = ln(AB)

klnA = ln(A^K)
5. ok ive worked out the first part - about the -ln6 + ln3 = 0.5 , just had to remember some of those rules, but still stuck on the ln0.5 = -ln2
6. lnA - lnB = ln(A/B)
lnA+lnB = ln(AB)

klnA = ln(A^K)
oh thanks - just worked that out !
7. (Original post by franks)
oh thanks - just worked that out !
If you remember the lna + lnb one, you can work out the klna one.

e.g. 3lna = lna + lna + lna = ln(aaa) = lna^3
8. thanks everyone, but can anyone please explain to me how ln0.5 = -ln 2 . Im probably missing a really simple point, but I just need someone to explain quickly!
9. (Original post by franks)
thanks everyone, but can anyone please explain to me how ln0.5 = -ln 2 . Im probably missing a really simple point, but I just need someone to explain quickly!
(-1)*ln2 = ln (2-1) = ln(½)

[log law: lnxn = nlnx]

Aitch
10. (Original post by Aitch)
(-1)*ln2 = ln (2-1) = ln(½)

[log law: lnxn = nlnx]

Aitch
you can also use another rule.
ln (0.5) = ln(1/2) = ln1 - ln2 = 0 - ln2 = -ln2

(btw, ln1 is always 0, as e^0=1)

[rule used = lnA - lnB = ln(A/B)]
11. thanks

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