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# core 3 paper question watch

1. show that integral of 2/x with limits root 6 and root 2 = ln3
2. (Original post by elvss567)
show that integral of 2/x with limits root 6 and root 2 = ln3
Go on then.
3. 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.
4. (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?
5. (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.
^^
6. (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

Similarly, try simplifying
7. (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.
8. Thankyou both!

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