Question 1176683
<pre>

The first step is to note that the lower limit of integration will be x=0.  This is because y=0 is one of the boundaries and y is negative for x<0.  So we will have limits of integration of x=0 and x=2.

Use integration by parts:

   Let   {{{u = x}}}       and    {{{ dv = e^(-x)dx }}}
giving us  {{{ du = dx}}}  and     {{{ v = -e^(-x) }}}


Recall integration by parts:
{{{ int(u dv) }}} = {{{u*v}}} - {{{int(v du)}}}

{{{ int(x e^(-x)dx) }}} = {{{x * (-e^(-x)) }}} - {{{int(-e^(-x) dx)}}}
                        = {{{ -x*e^(-x)) }}} - {{{e^(-x)}}}
                        = {{{ -e^(-x)(x+1) }}}

Evaluate this expression at x=2:  -0.4060  (approx)
                     and at x=0:   -1

Subtract the bottom from the top:  -0.4060-(-1) = 0.5940 sq units. (to 4 decimal places)