Let f and g be the functions given by #f(x)=e^x# and #g(x)=1/x#. What is the area of the region enclosed by the graphs of f and g between x=1 and x=2?

1 Answer
Apr 15, 2016

Answer:

#A=e^2-e-ln2# units

Explanation:

If we take a peek at the graphs of these functions, we'll see this:
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Of course, we won't see that monstrous yellow blob - that's me. But what we need to concentrate on are the red graph (#g(x)=1/x#) and the blue graph (#f(x)=e^x#). The area between these two functions is the yellow region - and that's what we're calculating.

To find it, all we need to do is find the area under #1/x# and subtract it from the area under #e^x#. We do this because the yellow region is really the area under #e^x#, without the area under #1/x#. Basically, take the area under #1/x# away and you have the yellow region.

Mathematically, we express the area of the region like this:
#A=int_1^2f(x)dx-int_1^2g(x)dx#

Because #f(x)=e^x# and #g(x)=1/x#,
#A=int_1^2e^xdx-int_1^2 1/xdx#

Lucky for us, these two integrals are quite simple. Because the derivative of #e^x# is #e^x#, the antiderivative is #e^x#; and because the derivative of #lnx# is #1/x#, the antiderivative is #lnx#:
#A=[e^x]_1^2-[lnx]_1^2#

Now it's on to evaluation:
#A=(e^(2)-e^(1))-(ln2-ln1)#
#A=e^2-e-ln2# units

An approximation to three decimal places is #A~~3.978# units.