What is the area bounded by y = e^x , x = 0, and y = 4e^-x?

Question

Other students have come up with the following answers, which is correct?

3 - 2(2^1/2 )

4 - 2(3^1/2 )

1

6 - 2(5^1/2 )

7 - 2(6^1/2 )

Answer 1

They intersect at (ln(2), 2)

The area below the upper curve but above the x axis is 2, below the lower is 1, so 2 - 1 = 1

1 = answer

Answer 2

For one thing, you must first determine where the two functions cross, in order to get the boundaries for integration. To find out where they intersect, you need to equate them to each other.

e^x = 4e^(-x)

e^x - 4e^(-x) = 0

e^(-x) [e^(2x) - 4] = 0

Note that e^(2x) is the same as (e^x)^2, so

e^(-x) [ (e^x)^2 - 4] = 0

Now, we have a difference of squares, so we can factor appropriately.

e^(-x) [e^x - 2] [e^x + 2] = 0

Therefore, e^(-x) = 0, e^x - 2 = 0, e^x + 2 = 0

e^(-x) = 0 cannot happen, and neither can e^x + 2 = 0. However, if e^x = 2, then x = ln(2).

So your boundaries of integration are from 0 to ln(2).

Now, you have to determine which function is "higher" on the graph between 0 and ln(2). Whichever function is the higher one will be the first function to subtract areas from. All we have to do is merely test a value between 0 and ln(2) for both functions. Let's use x = 0.5, and we find out using a calculator that the bigger function is 4e^(-x). So our area is calculated as follows.

A = Integral (4e^(-x))dx - Integral (e^x)dx [with both integral signs going from 0 to ln(2)]

Combine into one integral:

A = Integral (4e^(-x) - e^x) dx [still evaluated at 0 to ln(x)]

And then take the integral individually.

A = [-4e^(-x) - e^x] evaluated from 0 to ln(2)

A = [-4e^(-ln2) - e^(ln(2)] - (-4e^(0) - e^0)

Note: e and ln are inverses of each other, and if you take e to the power of ln(x), it just becomes x. This is no different; e^(ln2) = 2.

A = [-4/[e^(ln2)] - 2] - (-4 - 1)
A = [-4/2 - 2] - (-5)
A = [-2 - 2] + 5
A = -4 + 5
A = 1

Answer 3

Mathematically solve for the intersection of e^x and 4e^-x
e^x = 4e^-x
e^2x = 4
2x = ln 4
x = 1/2 ln 4 = ln4^1/2 = ln 2

next integrate 4e^-x - e^x (since subtracting their areas will leave the intersection) from zero to ln 2

I broke it into 2 parts:
integral 4e^-x = -4e^-x evaluated 0 -> ln 2
= -4 (e^-ln2 - e^0) = -4 (e^ln1/2 - e^0) = -4(1/2 - 1) = 2

integral e^x = e^x evaluated 0 -> ln 2
= e^ln2 - e^0 = 2 -1 = 1

integral 4e^-x - e^x from 0 -> ln2 = 2-1 = 1

Answer 4

use graphmatica, an online program which allows you to key in graphs to solve your problem.