Question

Sorry about all these integral equations...\[\int\limits_{0}^{In2}2e^{1-2x}dx\]

Answer 1

-e^(1-2x)
after limits
e^(1) - e^(1 - ln4) = e - 1/4. probably

Answer 2

The answer is \[{3\over4 }e\]

Answer 3

But my integral is right, maybe I did some mistake while applying limits

Answer 4

Oh wait, it's supposed to be e/4 not 1/4

Answer 5

\[-e^{1-2x}\] is integral, yes, I got that too :) It's the In I can't handle..

Answer 6

e^(1-ln4) =e^(lne -ln4) =e^(lne/4) = e/4
e - e/4 = 3/4e

Answer 7

How did you get that? I don't know how to handle the IN

Answer 8

ln(a) - ln(b) = ln(a/b)
e^(lnx) = x

these are the two properties I used, try it again.

Answer 9

How did you get 1 to become Ine?

Answer 10

\[1 = \log_a a\]
Depending upon the situation I used a=e.

Answer 11

So that's the same as Ina?

Answer 12

No, \[\log_e e= ln e = 1\]

Answer 13

Yes, I was referring to that. Ok. That makes more sense now :)... Just one question. How did you get e-e/4 at the end? Where did the first e come from?

Answer 14

Oh umm \[-e^{1-2x} \Large|_{0}^{\ln 2} \] \[-e^{1-2 \ln 2} + e^{1 -2 \times 0}}\]

Answer 15

Oh. Ok. thanks!