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core 3 paper question

show that integral of 2/x with limits root 6 and root 2 = ln3
Original post by elvss567
show that integral of 2/x with limits root 6 and root 2 = ln3


Go on then.
Quite a trivial problem once you: integral of 1/x = ln x (or absolute value for negative values).

Hence F(root6) - F(root 2) = ln3. You can show this now.
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elvss567
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Original post by RDKGames
Go on then.


i integrated 2/x to get 2lnx. then applying the limits gets me:

(2lnroot 6 ) - (2lnroot2) =

using log rules i got lnroot 3 as the answer.

the mark scheme answer says ln3. what have i done wrong when using log rules?
Reply 4
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elvss567
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Original post by thekidwhogames
Quite a trivial problem once you: integral of 1/x = ln x (or absolute value for negative values).

Hence F(root6) - F(root 2) = ln3. You can show this now.


^^
Original post by elvss567
i integrated 2/x to get 2lnx. then applying the limits gets me:

(2lnroot 6 ) - (2lnroot2) =

using log rules i got lnroot 3 as the answer.

the mark scheme answer says ln3. what have i done wrong when using log rules?


Well 2ln⁡(6)=ln⁡(6)=ln⁡(2)+ln⁡(3)2\ln (\sqrt{6})=\ln(6)=\ln(2)+\ln(3)

Similarly, try simplifying 2ln⁡(2)2\ln(\sqrt{2})
Original post by elvss567
i integrated 2/x to get 2lnx. then applying the limits gets me:

(2lnroot 6 ) - (2lnroot2) =

using log rules i got lnroot 3 as the answer.

the mark scheme answer says ln3. what have i done wrong when using log rules?


Use logarithm rules properly. 2ln root 6 = ln (root6)^2 = ln6
Similarly, the second part is ln2.

Thus ln6-ln2=ln3 as requested.
Reply 7
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elvss567
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Thankyou both!:smile:

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