Answer
$2.5(b-a) \ln \dfrac{d}{c}$
Work Step by Step
$J(u,v)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}$
or, $=\begin{vmatrix} 3.5u^{2.5}v^{-2.5}&-2.5v^{2.5}u^{-3.5}\\-2.5u^{3.5}u^{-3.5}& 2.5v^{1.5}u^{-2.5}\end{vmatrix}=\dfrac{2.5}{v}$
$\iint_R dA=\int_c^{d} \int_{a}^{b}\dfrac{2.5}{v} du dv=\int_c^d \dfrac{2.5}{v} dv \int_{a}^{b} du$
Therefore,
$\iint_R dA=2.5[\ln d -\ln c](b-a)=2.5(b-a) \ln \dfrac{d}{c}$
