Find the area of the region between y3e2x and ylnx for 1

Find the area of the region between y=3e^(-2x) and y=lnx for 1<x<2. Round your answer to three decimal places.

The functions meet at some point, call it z between 1 and 2 (about 1.268084)

3e^(-2x) is bigger up to z, then becomes smaller
The area is then

_1..z (3e^(-2x)-ln(x)) dx+_z..2 (ln(x)-3e^(-2x)) dx

e^(-2x)=-e^(-2x)/2

ln(x)=x(ln(x)-1) (you can use integration by parts u=ln(x), dv=1)

So we get

Area= -3e^(-2x)/2-x(ln(x)-1)|_1..z + x(ln(x)-1)+3e^(-2x)/2_z..2

Now plug in the numbers and the approximate value of z and you are done