Reduction formulae 

1) 

\[I_0=\int_0^1 (1-x)^0 e^{2x}dx=\int_0^1  e^{2x}dx=\frac{1}{2}e^{2x}|_0^{1}=\frac{e^2-1}{2}.\]

2) Let $u=(1-x)^n$ and $dv=e^{2x}dx$. Then 
\[du=-n(1-x)^{n-1},     v=\frac{1}{2}e^{2x}.\]
Integrating by parts we get 
$$I_n=\int_{0}^{1}e^{2x}(1-x)^ndx=(1-x)^n (\frac{1}{2}e^{2x})|_0^{1}-\int_0^{1} -n (1-x)^{n-1}(\frac{1}{2}e^{2x})dx$$
\[=0-\frac{1}{2}+\frac{n}{2}\int_0^1 (1-x)^{n-1}e^{2x}dx=\frac{n}{2}I_{n-1}-\frac{1}{2}.\]
So 
\[I_n=\frac{n}{2}I_{n-1}-\frac{1}{2}.\]
3) Using the above formula for $n=1$ and $n=2$ we get 
\[I_1=\frac{1}{2}I_{0}-\frac{1}{2}=\frac{1}{2}(\frac{e^2-1}{2})-\frac{1}{2}=\frac{e^2-1}{4}-\frac{1}{2}.\]
And
\[I_2=\frac{2}{2}I_{1}-\frac{1}{2}=I_1-\frac{1}{2}=\frac{1}{2}(\frac{e^2-1}{2})-\frac{1}{2}-\frac{1}{2}\]
\[=\frac{e^2-1}{4}-1.\]