anonymous
  • anonymous
e^.07(10-t)[10,1) find the definite integral
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
[10,1) is a little backwards... (1,10] is that e^.07 times (10-t)? or is the last part included in the exponent?
anonymous
  • anonymous
\[\int\limits_{1}^{10} e^.07(10-t)\]
amistre64
  • amistre64
so it aint included as the exponent..makes it alittle easier :)

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amistre64
  • amistre64
e^.07 is a constant, so pull it out of the way
amistre64
  • amistre64
(S) (10-t) dt -> 10t - (1/2)t^2 F(t) = [10t - (t^2)/2] e^.07
amistre64
  • amistre64
F(10) - F(1) will be the answer
amistre64
  • amistre64
(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 :)
amistre64
  • amistre64
that might be alittle off, did you mean to exclude 1 as an option?
anonymous
  • anonymous
no 1 is included. now I'm understanding
amistre64
  • amistre64
whew!!.... cause in the top {10,1) means everything from 10 to 1 but not including 1 :)
anonymous
  • anonymous
no i just want it for 10 and 1
amistre64
  • amistre64
then were good with it :) I got some idea for how to find it if it approaches 1, but nothing ti be sure about :)
anonymous
  • anonymous
Ive got another one its its kinda hard what you think
amistre64
  • amistre64
I can take a stab at it..... im ok with failure :)
amistre64
  • amistre64
is it one thats already posted? or you need to write it up?
anonymous
  • anonymous
im going to write it now
anonymous
  • anonymous
\[\int\limits_{0}^{15}\]e^0.05e^.06(15-t)
amistre64
  • amistre64
to clean it up... is that: e (e) (15-t) ??
amistre64
  • amistre64
or is that: e^e^(15-t)?
amistre64
  • amistre64
i assume the first one :)
amistre64
  • amistre64
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...
amistre64
  • amistre64
you can see the typo right....
amistre64
  • amistre64
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
anonymous
  • anonymous
\[\int\limits_{0}^{15}e^.05e.06(15-t)\]
anonymous
  • anonymous
this is how it looks exactly will u get the same answer or does this makes a difference
amistre64
  • amistre64
is that .06 an exponent?
amistre64
  • amistre64
(e^.05) (e^.06) (15-t) is what I integrated :)
amistre64
  • amistre64
that equation option down there is useful for some things, but I can never get it to do what I want....
amistre64
  • amistre64
\[\int\limits_{0}^{15} e^{.05} e^{.06} \left( 15-t \right)\]
amistre64
  • amistre64
\[= (112.5) e ^{.11}\]
anonymous
  • anonymous
it its e^0.5.e^.06 the 15-t is beside the .06 in exponential form
anonymous
  • anonymous
ok so what will happen to the ^(15-t)
amistre64
  • amistre64
\[\int\limits_{0}^{15} e ^{.05} e^{.06\left( 15-t \right)}\]
anonymous
  • anonymous
yes that's exactly what it is
amistre64
  • amistre64
ok....we can pull out that first e, its nothing but a constant
amistre64
  • amistre64
go ahead and distribute the .06 thru the (...)
amistre64
  • amistre64
e^(.8 - .06t) right?
amistre64
  • amistre64
(.9 - .06t)
amistre64
  • amistre64
If we can get this into the form: (S) Du e^u , then it suits up to e^u....does that make sense?
amistre64
  • amistre64
think back to derivatives... e^2x goes down to 2 e^2x right? does that ring a bell?
amistre64
  • amistre64
so... (S) 2 e^2x would equal just e^2x does that help out?
amistre64
  • amistre64
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 ??
amistre64
  • amistre64
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>
anonymous
  • anonymous
not sure im getting the last part
amistre64
  • amistre64
ok.... tell me what your having difficulites with and I can help iron out the wrinkles :)
anonymous
  • anonymous
ok the (.8-.06t) not getting the dirivitive of this
amistre64
  • amistre64
im thinking it involves the chain rule...which I think should be called the "gear"rule...
amistre64
  • amistre64
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?
anonymous
  • anonymous
yes i understand how you get the -.06 is this where we use the [15,0]
amistre64
  • amistre64
not yet, we havent gotten to our initial function yet, we still need to find a way to get there first.
amistre64
  • amistre64
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?
anonymous
  • anonymous
why did we substititute?
amistre64
  • amistre64
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.
amistre64
  • amistre64
what number do we know of that is used to obtain the same "value" but a different "shape" of a function? x times ? = x ??
amistre64
  • amistre64
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
anonymous
  • anonymous
ok
amistre64
  • amistre64
We have e^u we WANT Du e^u what value does Du have to be?
amistre64
  • amistre64
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 ?
amistre64
  • amistre64
think 1....think 1.... 1 times e^u = e^u right?
anonymous
  • anonymous
right
amistre64
  • amistre64
does Du/Du = 1?
amistre64
  • amistre64
does (-.06)/(-.06) = 1 ??
amistre64
  • amistre64
-.06 e^.05 (S) ----- e^(.8 -.06t) dt -.06 look close to what we want it to be?
amistre64
  • amistre64
if your unsure, tell me what your doubts are.... cause I can be rather stupid at times and mess alot of simple things up :)
amistre64
  • amistre64
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)]
amistre64
  • amistre64
or put another way: - e^(.85 - .06t) F(t) = ------------- .06
amistre64
  • amistre64
(Constants) [S] Du e^u du becomes (Constants) (e^u) is all we did in a nutshell
amistre64
  • amistre64
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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