Answer
$A = \frac{7}{4}\left( {{e^2} - 1} \right)$
Work Step by Step
$$\eqalign{
& y = \left( {3 + {x^2}} \right){e^{2x}},{\text{ from }}x = 0{\text{ to }}x = 1 \cr
& {\text{The area is given by}} \cr
& A = \int_0^1 {\left( {3 + {x^2}} \right){e^{2x}}} dx \cr
& {\text{Integrate by parts}} \cr
& u = 3 + {x^2},{\text{ }}du = 2xdx \cr
& du = {e^{2x}},{\text{ }}v = \frac{1}{2}{e^{2x}} \cr
& \int {\left( {3 + {x^2}} \right){e^{2x}}} dx = \frac{1}{2}\left( {3 + {x^2}} \right){e^{2x}} - \int {x{e^{2x}}dx} \cr
& {\text{Integrate by parts }}\int {x{e^{2x}}dx} \cr
& u = x,{\text{ }}du = dx \cr
& du = {e^{2x}},{\text{ }}v = \frac{1}{2}{e^{2x}} \cr
& \int {\left( {3 + {x^2}} \right){e^{2x}}} dx = \frac{1}{2}\left( {3 + {x^2}} \right){e^{2x}} - \underbrace {\int {x{e^{2x}}dx} }_ \downarrow \cr
& \int {\left( {3 + {x^2}} \right){e^{2x}}} dx = \frac{1}{2}\left( {3 + {x^2}} \right){e^{2x}} - \left( {\frac{1}{2}x{e^{2x}} - \int {\frac{1}{2}{e^{2x}}} dx} \right) \cr
& \int {\left( {3 + {x^2}} \right){e^{2x}}} dx = \frac{1}{2}\left( {3 + {x^2}} \right){e^{2x}} - \frac{1}{2}x{e^{2x}} + \frac{1}{2}\int {{e^{2x}}} dx \cr
& \int {\left( {3 + {x^2}} \right){e^{2x}}} dx = \frac{1}{2}\left( {3 + {x^2}} \right){e^{2x}} - \frac{1}{2}x{e^{2x}} + \frac{1}{4}{e^{2x}} \cr
& {\text{Therefore}}{\text{,}} \cr
& A = \int_0^1 {\left( {3 + {x^2}} \right){e^{2x}}} dx \cr
& A = \left[ {\frac{1}{2}\left( {3 + {x^2}} \right){e^{2x}} - \frac{1}{2}x{e^{2x}} + \frac{1}{4}{e^{2x}}} \right]_0^1 \cr
& A = \left[ {\frac{1}{2}\left( {3 + {{\left( 1 \right)}^2}} \right){e^2} - \frac{1}{2}{e^2} + \frac{1}{4}{e^2}} \right] - \left[ {\frac{1}{2}\left( 3 \right){e^0} - 0 + \frac{1}{4}} \right] \cr
& A = \frac{7}{4}{e^2} - \frac{7}{4} \cr
& A = \frac{7}{4}\left( {{e^2} - 1} \right) \cr
& A \approx 11.18 \cr} $$
