Find the area of the region bounded by #y=2e^x#, #y=e^(2x)# and #x=0#?

Question

#y=2e^x#, #y=e^(2x)# and #x=0#

16975 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-21T17:05:44

The area is #1/2# square units.

You will want to find the intersection points of the curve in order to correctly sketch the region.

#e^(2x) = 2e^x#

Let #e^x =t#.

#t^2 = 2t#

#t^2 - 2t = 0#

#t(t - 2) = 0#

#t = 0 or 2#

#e^x = 0 or e^x = 2#

#x = O/ or ln2#

Thus our interval will be #[0, ln2]#.

We now note that on #[0, ln2]#, the function #y = 2e^x# has a larger value than #y = e^(2x)#. Therefore, our integral will be

#A = int_0^(ln2) 2e^x -e^(2x)dx#

#A = [2e^x - 1/2e^(2x))]_0^(ln2)#

#A = 4 - 2 - (2 - 1/2(1))#

#A = 1/2# square units.

Hopefully this helps!