Question

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Integral of 0 to 6 (1/(x+2))dx = 2ln2

Answer 1

\[\int\limits_{0}^{6}(1/(x+2))dx = 2\ln2\]

Answer 2

\[\int\limits_{}^{} 1/(x+2) dx = \ln(x+2)\] (I am excluding the constant C since we are dealing with a definite integral)

So now we take the limit from 0 to 6 using the Fundamental Theorem of Calculus and get \[\ln 8 -\ln 2\].

This is the same as \[\ln 2^3 - \ln 2\]

Do you remember your properties for log functions?

Answer 3

Not too well. :/ I need to review them again.

Answer 4

Ok well \[\ln a^b = b*\ln a\]

Thus \[\ln 2^3 = 3\ln 2\]

so 3ln2 - ln 2 = 2ln2

Answer 5

ah ok. :D Thanks. Could you please explain to me how you got ln8 - ln2 at first though?

Answer 6

Suppose \[\int\limits_{}^{} fdx = F\]

The Fundamental Theorem of Calculus says \[\int\limits_{a}^{b} fdx = F(b) - F(a)\]

In this case \[1/(x+2) = f\] and \[\ln(x+2) = F\]

From there it's just plug n chug

Answer 7

Oh, and a = 0, b = 6 of course

Answer 8

Thank youuu