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

- anonymous

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

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- 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

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

- amistre64

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

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- amistre64

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

- amistre64

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

- amistre64

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

- 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

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

- anonymous

no 1 is included. now I'm understanding

- amistre64

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

- anonymous

no i just want it for 10 and 1

- 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

Ive got another one its its kinda hard what you think

- amistre64

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

- amistre64

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

- anonymous

im going to write it now

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

i assume the first one :)

- 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

you can see the typo right....

- 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

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

- anonymous

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

- amistre64

is that .06 an exponent?

- amistre64

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

- amistre64

that equation option down there is useful for some things, but I can never get it to do what I want....

- amistre64

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

- amistre64

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

- anonymous

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

- anonymous

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

- amistre64

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

- anonymous

yes that's exactly what it is

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

(.9 - .06t)

- 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

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

- amistre64

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

- 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

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

not sure im getting the last part

- amistre64

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

- anonymous

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

- amistre64

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

- 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

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

- amistre64

not yet, we havent gotten to our initial function yet, we still need to find a way to get there first.

- 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

why did we substititute?

- 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

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

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

ok

- amistre64

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

- 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

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

- anonymous

right

- amistre64

does Du/Du = 1?

- amistre64

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

- amistre64

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

- 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

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

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

- amistre64

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

- 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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