Question

Find the area bounded by the graphs of

Answer 1

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 ...

  $\displaystyle A=\int_1^3{\left(\dfrac{1}{2}\ln(y)-\dfrac{-1}{4}\ln(y)\right)}\,dy=\dfrac{3}{4}\int_1^3{\ln(y)}\,dy=\dfrac{3}{4}\left.(x\ln{(x)}-x)\right|_1^3\\\\=\dfrac{3}{4}(3\ln(3)-(3-1))=\boxed{\dfrac{9\ln(3)-6}{4}\approx0.97187765}$

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The attachment shows a numerical integration using a vertical slice of area. The result is identical to 12 significant figures.