The area of the region enclosed by the curve y=x³ and its tangent at the point (-1,-1) is 

$\mathrm{y} = {\mathrm{x}}^3$
$\frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{x}}^2$
${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{(-1,-1)}=3$
Equation of tangent at (-1,-1)  is y + 1 = 3(x+1)
$\Rightarrow 3\mathrm{x}-\mathrm{y}+2=0$
Solving y = x³; $3\mathrm{x}-\mathrm{y}+2=0$  for point of intersection, we get Q=(2,8)
Required area= ${\int }_{-1}^2\left[(3\mathrm{x}+2)-{\mathrm{x}}^3\right]\mathrm{dx}$
$={[\frac{3{\mathrm{x}}^2}{2}+2\mathrm{x}-\frac{{\mathrm{x}}^4}{4}]}_{-1}^2=(6+4-4)-(\frac{3}{2}-2-\frac{1}{4})$
$=6+\frac{3}{4}=\frac{27}{4}$ Sq. Units