Question

\[\int\limits_{1/2}^{2}({e^(1/x) \over x^2})dx\]

Answer 1

\[\int\limits_{1/2}^{2} {e^{1/x} \over x^2}dx\]

Answer 2

ignore that first bit

Answer 3

12

Answer 4

can you tell me how you got that?

Answer 5

Use a substitution \(u=\frac{1}{x}\).

Answer 6

ok, this is what I've tried:
\[\int\limits_{1/2}^{2}{1/x \over 2 \ln {x}} dx = \int\limits_{-\ln2}^{\ln2}{1 \over 2u}du = {1 \over 2}\ln |u|]_{-\ln2}^{\ln 2} = 0\]

Answer 7

that last step, btw, was
(1/2) (ln|ln 2| - ln|-ln 2|) = (1/2)(ln|ln 2| - ln|ln 2|) = 0

Answer 8

It seems like you've taken ln of the top and bottom and you're not allowed to do that, it's wrong. Note for example that \(\frac{1}{1}\ne \frac{\ln{1}}{\ln{1}}\). Try to use the substitution I wrote above, it will work.

Answer 9

ok, I'll give it a go

Answer 10

@everyone, is the equation editor still functionable? i think it's not,

Answer 11

oh, so you can take ln of equations but not expressions, is that what it was?

Answer 12

try to use the equation editor. . i think it's not working. . the tab beside draw tab

Answer 13

True, you can take ln of both sides of an equation. But you can't take ln of the top and bottom of an expression.

Answer 14

\[\text{It's working with me.}\]

Answer 15

no it's not with me. . i don't know what happened. .

Answer 16

\[Its working!!!\]

Answer 17

Refresh your page.

Answer 18

i can't even look unto equations

Answer 19

I got it :) \[e^2-\sqrt{e}\]
And my equation thing's working

Answer 20

i think it's okay now. . . thank you guys. .

Answer 21

You've got it right!

Answer 22

Thanks for the help!

Answer 23

Anytime.