Question

Definite Integrals

Answer 1

try doing a u sub
let u = ln(x^2) and work from there

Answer 2

Ok ill try that one second

Answer 3

I managed to get 4 as the answer, I don't know if it's right or not

Answer 4

can i see the work you did?

Answer 5

so you should have something like this
\[u = \ln(x^2)\]
\[\frac{ 1 }{ 2 } du = \frac{ dx }{ x }\]


\[\int\limits_{}^{} \frac{ 1 }{ 2 }u du\]

Answer 6

then you factor out the 1/2 and integrate that u

Answer 7

so when you integrate you get
\[\frac{ 1 }{ 4 } \ (ln(x^2))^2 \]
evaluated from 1 to e

Answer 8

so when you evaluate it looks like
\[\frac{ 1 }{ 4 } (\ln(e^2))^2 - \frac{ 1 }{ 4 } (\ln(1^2))^2\]
\[= \frac{ 1 }{ 4 } (\ln(e^2))^2 - 0\]
\[= \frac{ 1 }{ 4 } (\ln(e*e))^2\]
\[= \frac{ 1 }{ 4 } (\ln(e) + ln(e))^2\]
\[= \frac{ 1 }{ 4 } (\ 1 + 1)^2\]

Answer 9

I suggest from the beginning to make
\[
\frac{\ln(x^2)} x= \frac {2\ln(x)} x
\]
It would have made computation a bit easier

Answer 10

^^ yea that works too, its best to simiplfy the prob before trying to solve it

Answer 11

and to notice that the integral is
\[

\int \frac{2 \ln (x)}{x} \, dx=\ln ^2(x)
\]

Answer 12

Hmm so the answer is 1 ?

Answer 13

\[
\int_1^e \frac{2 \ln (x)}{x} \, dx=\ln^2 (e)-\ln^2 (1)=1
\]

Answer 14

correct

Answer 15

Well I got 1 my first time trying this problem, but when I checked it on a calculator, i think it was mathway, I got 1/2, I guess I typed it wrong lol

Answer 16

Thanks for the help!

Answer 17

your welcome