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\[\int\limits_{1}^{10} e^.07(10-t)\]

so it aint included as the exponent..makes it alittle easier :)

e^.07 is a constant, so pull it out of the way

(S) (10-t) dt -> 10t - (1/2)t^2
F(t) = [10t - (t^2)/2] e^.07

F(10) - F(1) will be the answer

(100 - 50)e^.07 - (10-.5)e^.07
(50 - 9.5) e^.07
(40.5) e^.07 if I did it right in me head :)

that might be alittle off, did you mean to exclude 1 as an option?

no 1 is included. now I'm understanding

whew!!.... cause in the top {10,1) means everything from 10 to 1 but not including 1 :)

no i just want it for 10 and 1

Ive got another one its its kinda hard what you think

I can take a stab at it..... im ok with failure :)

is it one thats already posted? or you need to write it up?

im going to write it now

\[\int\limits_{0}^{15}\]e^0.05e^.06(15-t)

to clean it up... is that:
e (e) (15-t) ??

or is that:
e^e^(15-t)?

i assume the first one :)

you can see the typo right....

\[\int\limits_{0}^{15}e^.05e.06(15-t)\]

this is how it looks exactly will u get the same answer or does this makes a difference

is that .06 an exponent?

(e^.05) (e^.06) (15-t) is what I integrated :)

\[\int\limits_{0}^{15} e^{.05} e^{.06} \left( 15-t \right)\]

\[= (112.5) e ^{.11}\]

it its e^0.5.e^.06 the 15-t is beside the .06 in exponential form

ok so what will happen to the ^(15-t)

\[\int\limits_{0}^{15} e ^{.05} e^{.06\left( 15-t \right)}\]

yes that's exactly what it is

ok....we can pull out that first e, its nothing but a constant

go ahead and distribute the .06 thru the (...)

e^(.8 - .06t) right?

(.9 - .06t)

If we can get this into the form:
(S) Du e^u , then it suits up to e^u....does that make sense?

think back to derivatives...
e^2x goes down to 2 e^2x right? does that ring a bell?

so...
(S) 2 e^2x would equal just e^2x does that help out?

not sure im getting the last part

ok.... tell me what your having difficulites with and I can help iron out the wrinkles :)

ok the (.8-.06t) not getting the dirivitive of this

im thinking it involves the chain rule...which I think should be called the "gear"rule...

yes i understand how you get the -.06 is this where we use the [15,0]

why did we substititute?

ok

We have e^u we WANT Du e^u
what value does Du have to be?

think 1....think 1.... 1 times e^u = e^u right?

right

does Du/Du = 1?

does (-.06)/(-.06) = 1 ??

-.06
e^.05 (S) ----- e^(.8 -.06t) dt
-.06
look close to what we want it to be?

or put another way:
- e^(.85 - .06t)
F(t) = -------------
.06

(Constants) [S] Du e^u du becomes
(Constants) (e^u) is all we did in a nutshell