Using integration, it is found that the area under the graph of f(x) in the desired interval is of 1 unit squared.
The area under the graph of a curve f(x) between x = a and x = b is given by the following integral:
[tex]A = \int_{a}^{b} f(x) dx[/tex]
In this problem, the function and the limits of integration are given by:
[tex]f(x) = \frac{1}{x\ln{x}}[/tex]
[tex]a = e, b = e^e[/tex]
Hence:
[tex]A = \int_{e}^{e^e} \frac{1}{x\ln{x}} dx[/tex]
Using substitution:
[tex]u = \ln{x}[/tex]
[tex]du = \frac{1}{x} dx[/tex]
[tex]dx = x du[/tex]
Hence:
[tex]\int \frac{1}{x\ln{x}} dx = \int \frac{1}{u} du = \ln{u} = \ln{\ln{x}}[/tex]
Applying the Fundamental Theorem of Calculus:
[tex]A = \ln{\ln{e^e}} - \ln{\ln{e}} = \ln{e\ln{e}} = \ln{\ln{e}} = \ln{e} - \ln{1} = 1[/tex]
Hence, the area under the graph of f(x) in the desired interval is of 1 unit squared.
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