Question 1085535
{{{ graph(400,400, -2,2, -2,2, x^3, x)  }}}
—

Noting that the curves cross at x=0 and x=1, with  {{{x>=x^3 }}} on that interval, and by symmetry, the total area bounded is twice the area of the bounded region in the first quadrant,  we can integrate from 0 to 1 and then just double it.  

{{{ Area = 2 * int( (x-x^3), dx, 0, 1 ) }}}
{{{ Area = 2* ((1/2)x^2 - (1/4)x^4) }}}  <<< evaluated at 1 and 0
{{{ Area = 2 *(1/2 - 1/4 - (0 - 0)) = 2(1/4) = highlight(1/2) }}}
======================

For v(t) = &#8722;t^2 + 6, the displacement on t=0 to t=6 is:

{{{ int((-t^2 + 6), dt, 0, 6) }}}
= {{{  -(1/3)t^3 + 6t }}}  evaluated at t=6 and t=0
= {{{ (-6^3)/3 + 6*6 - (-0 + 0) }}} 
= {{{ -72 + 36 = highlight_green(-36ft) }}}

The interpretation of -36ft means the displacement is "to the left" assuming you are using the usual convention of positive displacement/velocity/accelerations  toward the right, along the positive x-axis.