Answer
$2.5(b-a) \ln \dfrac{d}{c}$
Work Step by Step
$Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} 3.5u^{2.5}v^{-2.5}&-2.5v^{2.5}u^{-3.5}\\-2.5u^{3.5}u^{-3.5}& 2.5v^{1.5}u^{-2.5}\end{vmatrix}=2.5v^{-1}$
Now, we have $\iint_R dA=\int_c^{d} \int_{a}^{b}\dfrac{2.5}{v} du dv$
$\implies \int_c^d 2.5v^{-1} dv \int_{a}^{b} du$
$\implies 2.5(b-a)[\ln d -\ln c]$
Hence, we have $\iint_R dA=2.5(b-a) \ln \dfrac{d}{c}$
