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Wiley PL-i C secure https// willey plus.com/edugen/student/mainfr uni edugen Apps n Manrile at llMass Am D SAIRFI ngon D Wiley Plias Homewnri Alpi Cnmr M Mathway l Math P P Toggers n PH 131 P Sign In I Mastering Q 20 bones Flashcards MasteringARP Contact Us Wiley PLUS Hughes-Hallett, Applied Calculus, 4e APPI TFD CAI CUI US (MATH 27/128 ASSIGNMENT RFsouRoPS Dunng a surge in the demand for electricity, the rate, r, at which energy is used can be approximated by chapter z Sectlon 4 where t s the time in hours and a s a positive constant. pter 7, Sect Qurstinn 15 (a) Find the total energy, E used in the first hours. Give your answer as a function af a Rcui Kesults by study objective (b) What happens to E as T- oc Click here to enter or edit your answer a O Ask me any 3/1/2017

Explanation / Answer

Integration by parts

int(u)dv = uv - int(v)du

u = t
du = dt

dv = e^(-at)dt
v = -(e^(-at)/a)

so int(u)dv = (-t/a)(e^(-at)) - (-a^-1)int(e^(-at))dt

= (-t/a)(e^(-at)) + (1/a)(-e^(-at)/a) + C
=(-t/a)(e^(-at))-(1/a^2)(e^(-at)) + C
= -(e^(-at))(at+1)/(a^2) + C

Note it's best to do this on paper rather then straight through in your head, these type of problems tends to get confusing.

a.)

So E = int0->T(r(t))dt
or

-(e^(-at))(at+1)/(a^2)[T->0]

-(e^(-aT))(aT+1)/(a^2) + (1/a^2)

b.) Basically we first check if E is convergent or divergent.

lim(e^(-aT) as T approaches inf evaluates to just 0
lim(aT+1)/(a^2) as T approaches inf evaluates to inf

now let's simplify both equations since we have a 0*oo limit so we can analyze the rate of convergence and divergence of this function (Note the only viable options now are either 0 or divergent)

we now assume that the following limits have T going to inf

lim(aT+1)/(a^2) =~ lim(T)
lim(e^(-aT) =~ lim(e^-T)

we know for a fact that lim(3^-T) has a much higher magnitude rate that lim(T) so the function converges at 0. Thus E approaches 0 as T approaches Inf on the right term.

Edited ----------------------------------------... (Forgot what E really was at first :S)

E = -(e^(-aT))(aT+1)/(a^2) + (1/a^2)

So in this case, the Real E evaluates to (1/a^2) since -(e^(-aT))(aT+1)/(a^2) approaches 0

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