Question

The value of integral \(\mathop \smallint \limits_0^2 \mathop \smallint \limits_0^{\rm{\ \ \ \ x}} {{\rm{e}}^{{\rm{x}} + {\rm{y}}}}{\rm{dydx}}\) is
1. \(\frac{1}{2}\left( {{\rm{e}} - 1} \right)\)
2. \( \frac{1}{2}{\left( {{{\rm{e}}^2} - 1} \right)^2}\)
3. \(\frac{1}{2}\left( {{{\rm{e}}^2} - {\rm{e}}} \right)\)
4. \(\frac{1}{2}{\left( {{\rm{e}} - \frac{1}{{\rm{e}}}} \right)^2}\)

Answer 1

Correct Answer - Option 2 : \( \frac{1}{2}{\left( {{{\rm{e}}^2} - 1} \right)^2}\)

\(\mathop \smallint \limits_0^2 \mathop \smallint \limits_0^{\rm{x}} {{\rm{e}}^{{\rm{x}} + {\rm{y}}}}{\rm{dydx}} = \mathop \smallint \limits_0^2 {{\rm{e}}^{\rm{x}}}\left( {\mathop \smallint \limits_0^{\rm{x}} {{\rm{e}}^{\rm{y}}}{\rm{dy}}} \right){\rm{dx}}\)

\(= \mathop \smallint \limits_0^2 {{\rm{e}}^{\rm{x}}}\left( {{{\rm{e}}^{\rm{y}}}} \right)_0^{\rm{x}}{\rm{dx}} = \mathop \smallint \limits_0^2 {{\rm{e}}^{\rm{x}}}\left( {{{\rm{e}}^{\rm{x}}} - 1} \right){\rm{dx}}\)

\(= \mathop \smallint \limits_0^2 \left( {{{\rm{e}}^{2{\rm{x}}}}-{{\rm{e}}^{\rm{x}}}} \right){\rm{dx}} = \left( {\frac{{{{\rm{e}}^{2{\rm{x}}}}}}{2} - {{\rm{e}}^{\rm{x}}}} \right)_0^2\)

\(= \frac{{{{\rm{e}}^4}}}{2} - {{\rm{e}}^2} - \frac{1}{2} + 1 = \frac{{{{\rm{e}}^4}}}{2} - {{\rm{e}}^2} + \frac{1}{2}\)

\(= \frac{1}{2}\left( {{{\rm{e}}^4} - 2{{\rm{e}}^2} + 1} \right) = \frac{1}{2}{\left( {{{\rm{e}}^2} - 1} \right)^2}\)