Question

integral 1/x dx
upper limit 5e lower limit 1

Answer 1

anti derivative is ?

Answer 2

natural log

Answer 3

evaluate the integral
\[\int\limits_{1}^{5e} \frac{ 1 }{ x } dx\]

Answer 4

do you know what the anti derivative of \(\frac{1}{x}\) is? which is another way of asking
"what would you take the derivative of to get \(\frac{1}{x}\) ? "

Answer 5

is it ln\[\left| x \right|\]

Answer 6

yes

Answer 7

so your last job is to plug in \(e^5\) and then plug in \(1\) then subtract

Answer 8

i.e. compute \(\ln(e^5)-\ln(1)\)

Answer 9

she threw me off because it says 5e not e^5, but i don't know how to find the derivative of 5e.

Answer 10

oh wait, it was not \(e^5\) it was \(5e\) right?

Answer 11

oh now you do not have to find the derivative of anything

Answer 12

you only have to compute the number

Answer 13

\[f(x)=\frac{1}{x}\]
\[F(x)=\ln(x)\]
\[\int_1^{5e}f(x)dx=F(5e)-f(1)=\ln(5e)-\ln(1)\]

Answer 14

final answer would be ln(5e)-0 = ln(5e) since ln(1) is 0

Answer 15

your only job now is to compute
\[\ln(5e)-\ln(1)\] then you are done

Answer 16

so i don't have to use the rule \[\ln \frac{ 1 }{ x} = \frac{ u' }{ u }\]

Answer 17

\(\ln(1)=0\) so you are left with \(\ln(5e)\) which guess you could rewrite as
\[\ln(5e)=\ln(5)+\ln(e)=\ln(5)+1\]

Answer 18

no not at all
you do not have \(\frac{u'}{u}\) you just have \(\frac{1}{x}\)

Answer 19

at any rate there is no such rule as \[\ln \frac{ 1 }{ x} = \frac{ u' }{ u }\]

Answer 20

what i think you mean is that
\[\frac{d}{dx}\ln(u)=\frac{u'}{u}\]

Answer 21

the 1 in ln(5e)+1 is just a constant. can it be completely ignored @satellite73?

Answer 22

@galacticwavesXX the whole thing, is a constant (since ln(5e) is a constant too), and we can't ignore any of it (but the ln1 is equal to zero)

\[\Large \int\limits\limits\limits_{1}^{5e} \frac{ 1 }{ x } dx = \left[ \ln x \right]^{5e}_{1} = \ln (5e) - \ln 1\]