Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. (Round your answer to four decimal places.)


Enclosed by y = ln x, y = 2 − ln x, and x = 4

you will need the intersection of

y = lnx and y = 2-lnx

2lnx= 2
lnx = 1
x = e

so you will have a region shaped like an "arrow head" from x=e to 4

Area = ∫(2 - 2lnx) dx from x = e to 4
(remember the integral of lnx is
xlnx - x , which should be in your repertoire of common integrals )

= [-2xlnx + 4x] from e to 4
= -8ln4 + 16 - (-2e + 4e)
= .....

To find the area of the indicated region enclosed by the curves y = ln(x), y = 2 - ln(x), and x = 4, we need to find the points of intersection between these curves and then integrate the functions to get the area.

1. Start by graphing the two curves: y = ln(x) and y = 2 - ln(x). You can use any graphing software or a graphing calculator for this step.

2. By analyzing the graph, we can see that the two curves intersect at a certain point. Let's find this point.

Set ln(x) equal to 2 - ln(x):
ln(x) = 2 - ln(x)

Adding ln(x) to both sides:
2ln(x) = 2

Dividing both sides by 2:
ln(x) = 1

Exponentiating both sides with base e:
x = e^1
x = e

Therefore, the curves intersect at the point (e, 1).

3. Now that we have the points of intersection, we need to determine the range of x-values that we are interested in. In our case, we are interested in the region up to x = 4.

4. To find the area, we will integrate the difference between the two curves with respect to x from x = e to x = 4:

A = ∫[e to 4] [(2 - ln(x)) - ln(x)] dx

5. Evaluate the integral:

A = ∫[e to 4] (2 - 2ln(x)) dx

A = [2x - 2xln(x)] evaluated over the range [e to 4]

A = (2(4) - 2(4)ln(4)) - (2(e) - 2(e)ln(e))

A = (8 - 8ln(4)) - (2e - 2e) [Note: ln(e) = 1]

Simplifying further:

A = 8 - 8ln(4) - 2e + 2e

A = 8 - 2e - 8ln(4)

6. Calculate the numerical value of the area using a calculator or software. Round the answer to four decimal places.