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?

Question

Evelyn Agard · Accepted Answer

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

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#.

Emily Ammerman · Answer

undefined To find the area of the region $ R $ enclosed by the graph of $ y = 2e^{-x} $ and the line $ x = k $ in the first quadrant, you need to integrate the function $ 2e^{-x} $ with respect to $ x $ over the interval from $ x = 0 $ to $ x = k $. The area of $ R $ in terms of $ k $ can be expressed as follows:

$$ \text{Area of } R = \int_{0}^{k} 2e^{-x} \, dx $$