e^.07(10-t)[10,1) find the definite integral

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e^.07(10-t)[10,1) find the definite integral

Mathematics
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[10,1) is a little backwards... (1,10] is that e^.07 times (10-t)? or is the last part included in the exponent?
\[\int\limits_{1}^{10} e^.07(10-t)\]
so it aint included as the exponent..makes it alittle easier :)

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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
then were good with it :) I got some idea for how to find it if it approaches 1, but nothing ti be sure about :)
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 :)
recall that like bases when multiplies add exponents, for example: 5^3 5^5 = 5^(3+5) = 5^8 so, e^.06 e^.06 = e^.11 which is still a constant and can be pulled out...
you can see the typo right....
that leaves us with: (S) 15-t dt -> 15t - (t^2)/2 | [15,0] F(t) = [15t - (t^2)/2] e^.11 since F(0) = 0 the only important on is F(15) F(15) = [15(15) - 15(15)/2] e^.11 = [225 - 112.5] e^.11 F(15)= 112.5 e^.11
\[\int\limits_{0}^{15}e^.05e.06(15-t)\]
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 :)
that equation option down there is useful for some things, but I can never get it to do what I want....
\[\int\limits_{0}^{15} e^{.05} e^{.06} \left( 15-t \right)\]
\[= (112.5) e ^{.11}\]
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?
our exponent here (.8-.06t) derives to: -.06 right? so we need to modify this set up to include an (-.06) without actually changing the value of the set up... what number can we multiply ANY number by to get the same value back? x times ? equals x ??
d (e^x) dx --- = ---- e^x dx dx Or to right it another way: Dx(e^x) -> Dx e^x Dx(e^7x) -> Dx(7x) e^7x -> 7 e^7x right>
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...
I want you to think of a box of gears that are all meshed together so that when you turn the very first one, it has an effect on all the rest of them. each gear in turn is turning another correct?
yes i understand how you get the -.06 is this where we use the [15,0]
not yet, we havent gotten to our initial function yet, we still need to find a way to get there first.
we have a function: e^u that depends on "u" for its value right? u = .8 -.06t and so the value of "u" depends on the value of "t" right?
why did we substititute?
we substitued values to see how the original function behaves. What we need in order to integrate this original function is to modify the way it looks without changing its "value". we can easily integrate the function Du e^u to get e^u so this is our goal. We need to modify the "shape" of our original function so that it matches the "easy to integrate" function without changing the "value" of our original function.
what number do we know of that is used to obtain the same "value" but a different "shape" of a function? x times ? = x ??
for example: say you only have a 20 dollar bill, and you want to buy some bubble gum for 1.00 but the cashier has no change to plit your $20. How can you purchase the bubble gum? by changing your $20 into a bunch of $1 bills... that is all we are wanting to do here.... so that we can make life easier for us
ok
We have e^u we WANT Du e^u what value does Du have to be?
Du is probably a bad name for that.... let me ask this.. what do we have to multiply (e^u) by in order to keep the same "value" e^u times ____ = e^u ?
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?
if your unsure, tell me what your doubts are.... cause I can be rather stupid at times and mess alot of simple things up :)
in essense, we want to slide that bottom (-.06) out of the way since it is a constant we can do that and put it under the left side. that leaves the integrand to be -(e^.05)/.06 [S] -.06 e^(.8-.06t) dt -> e^(.8-.06t) F(t) = - [(e^.05)/.06] [e^(.8-.06t)]
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
I gotta head to class for the next few hours... if your still lost, go ahead and post it for everyone to see :) Ciao

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