Let R be the first quadrant region enclosed by the graph of #y= 2e^-x# and the line x=k, how od you find the area of R in terms of k?

1 Answer
Apr 30, 2016

Answer:

Integrate #y# from #x=0# to #x=k# to get #R=2-2e^(-k)#.

Explanation:

We know that the area under a curve can be found by integrating it; so the area under #y=2e^(-x)# is given by: #int2e^(-x)dx#. The only problem is what #x#-values we need to integrate.

The first quadrant starts at #x=0#, and continues for all positive values of #x#. The problem states that we're finding the area under #y# in the first quadrant (which begins at #x=0#) to the line #x=k#. So we'll be integrating from #x=0# to #x=k#:
#R=int_0^k2e^(-x)dx#
#R=2int_0^ke^(-x)dx#
#R=2[-e^(-x)]_0^k#
#R=2(-e^((-k))-(-e^((-0)))#
#R=2(-e^(-k)+1)=2-2e^(-k)#

Note that #2e^(-k)# will never be greater than #2#, which means the area will always be greater than #0#.