and 
in the first quadrant is the set
, so that 
. Then when 
, you also have 
; when 
. Then the integral is equivalent to
![\bf \displaystyle\int\limits_{0}^{2}~e^{\frac{x}{2}}\cdot dx \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using some substitution}}{u=\cfrac{x}{2}\implies u=\cfrac{1}{2}x\implies} \cfrac{du}{dx}=\cfrac{1}{2}\implies 2du=dx \\\\[-0.35em] ~\dotfill\\\\ \displaystyle\int\limits_{0}^{2}~e^{u}\cdot 2du\implies 2\int\limits_{0}^{2}~e^u\cdot du \\\\[-0.35em] ~\dotfill](/tpl/images/0013/5416/885bf.png)
![\bf \stackrel{\textit{changing the bounds}}{u(0)=\cfrac{0}{2}}\implies u(0)=\boxed{0}~\hspace{7em} u(2)=\cfrac{2}{2}\implies u(2)=\boxed{1} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle 2\int\limits_{0}^{1}~e^u\cdot du\implies \left. 2e^u\cfrac{}{} \right]_{0}^{1}\implies 2e-2](/tpl/images/0013/5416/affd9.png)
Ignore the chicken but how do you do #10
...
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