Answer:
(9ln(3) -6)/4 ≈ 0.97187765
Step-by-step explanation:
A horizontal slice of the area will range from g^-1(y) to f^-1(y). The inverse functions are ...
f^-1(x) = 1/2·ln(x)
g^-1(x) = -1/4·ln(x)
Then the integral can be written as ...
[tex]\displaystyle A=\int_1^3{\left(\dfrac{1}{2}\ln(y)-\dfrac{-1}{4}\ln(y)\right)}\,dy=\dfrac{3}{4}\int_1^3{\ln(y)}\,dy=\dfrac{3}{4}\left.(x\ln{(x)}-x)\right|_1^3\\\\=\dfrac{3}{4}(3\ln(3)-(3-1))=\boxed{\dfrac{9\ln(3)-6}{4}\approx0.97187765}[/tex]
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The attachment shows a numerical integration using a vertical slice of area. The result is identical to 12 significant figures.
