Question

Let R be the region between the graph of y=e^-2x and the x-axis for x greater than or = to 3. The area of R is?

Answer 1

are you familiar with integrals?

Answer 2

yes.

Answer 3

its just i dont now what my upper and lower limit should be.

Answer 4

and im bad at integrating

Answer 5

Since x is greater than or equal to 3, this means that the lower limit is 3


The upper limit will be infinity. This is valid because e^(-2x) is a decreasing function (it will slowly approach 0)

Answer 6

So this problem deals with improper integrals. Have you had experience working with improper integrals?

Answer 7

oh yeah i did

Answer 8

i dont really remeber it though

Answer 9

int( e^(-2x) dx) .... Note: this means "integral of e^(-2x) with respect to x"


Let u = -2x, so du/dx = -2 ---> du = -2dx ---> dx = -du/2


So int( e^(-2x) dx) turns into int( e^u (-du/2))


Rearrange terms to get


-1/2*int( e^u du)


and then integrate


-1/2*(e^u + C)


-1/2*e^u + C


-1/2*e^(-2x) + C


So the integral of e^(-2x) is -1/2*e^(-2x) + C


With me so far?

Answer 10

ok lemme just read it..give me a minute

Answer 11

sure thing

Answer 12

yupp i get it so far

Answer 13

that's great...my apologies, I got distracted


Now evaluate the integral at its limits


In general, if the limits of this integral were from 'a' to 'b', then


int(e^(-2x) dx) = (-1/2e^(-2b) + C) - (-1/2e^(-2a) + C)


int(e^(-2x) dx) = -1/2e^(-2b) + 1/2e^(-2a)


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In our case, a = 3 and b = t where t approaches infinity, so...


int(e^(-2x) dx) = -1/2e^(-2t) + e^(-2*3)


int(e^(-2x) dx) = -1/2e^(-2t) + 1/2e^(-6)



From here, you take the limit as t goes to infinity. This will make -1/2e^(-2t) approach 0. This effectively makes the first term -1/2e^(-2t) go away leaving us with 1/2e^(-6)


This means that the answer is 1/2e^(-6) which can be written as 1/(2e^6)


Hopefully this makes sense. If not, let me know. Thanks.

Answer 14

So u know how u subtract the two values u get when u subsitute the upper and lower limit. When u substitute a 3 u get 1/2e^(-6) and when u subtistute a t it just goes to 0. So the asnwer is just 1/2e^(-6). Am i understanding it correctly?

Answer 15

when I substitute t and let t approach infinity, the first term -1/2e^(-2t) goes to zero, so yes you have the right idea.

Answer 16

THANK YOU SO MUCH! AND TAKING THE TIME TO EXPLAIN THOROUGHLY. This was extremely helpful. =).