Question

another u substitution problem, i dont know what to use as u.

Answer 1

\[\int\limits_{0}^{1} \frac{ 2e ^{6x} -5}{ e ^{2x} }dx\]

Answer 2

at first glance i would say divide each term by \(e^{2x}\)

Answer 3

gives you \[2\int_0^1e^{4x}dx-5\int e^{-2x}dx\]

Answer 4

i forgot the limits on the second one, but you get the idea right?

Answer 5

yes,i'll try it out

Answer 6

would i only use u substitution on the second half?

Answer 7

both are u-sub problems, but they are really mental u subs, i.e. do them in your head

Answer 8

fist is \(u=4x\) second is \(u=-2x\)

Answer 9

*first

Answer 10

\[\frac{ 1 }{ 4 }\times2 \int\limits_{0}^{4}e ^{u}du-5\times2\int\limits_{0}^{2}\frac{ 1 }{ e ^{u} }du\]

Answer 11

how would i do the integral of e^u if u is 4x and 2x? would it be e^u/ u ?

Answer 12

the anti derivative of \(e^u\) is \(e^u\)
it is it's own father

Answer 13

but it would save you effort if you made the second sub for \[-5\int_0^1e^{-2x}dx\] as \[u=-2x\]

Answer 14

i think you also got confused, because the first one you did correctly
\[u=4x\\
du=4dx\\
\frac{1}{4}du=dx\] but in the second one you multiplied instead of dividing

Answer 15

try \[-5\int_0^1e^{-2x}dx\] making \(u=-2x\) and see what you get

Answer 16

\[\frac{ 1 }{ 2 }(e ^{16}-e^0) - \frac{ 5 }{ 2} (e^0-e^4)\]

Answer 17

its wrong :(

Answer 18

ok there are a couple mistakes here but we can fix them

Answer 19

first integral you changed the limits of integration when you made the u - sub right?

Answer 20

you got \[\frac{1}{2}\int _0^4e^udu\] which is correct
but since you changed the limits of integration you do not have to change back
the anti derivative of \(e^u\) is \(e^u\) so first integral is \[\frac{1}{2}(e^4-e^0)\]

Answer 21

i think you changed back to \(e^{4x}\) but then replaced \(x\) by 4
if you change back, then you would replace x by 1 (and get 4)
changing the limits of integration allows you not to switch back

Answer 22

clear, ot no?

Answer 23

*or

Answer 24

ok, i got it, thank you!