The area of the region enclosed by the curve y = x 3 and its tangent at the point ( – 1 ,  …

Given $y=x^3 ... \left(\mathrm{i}\right)$

$\Rightarrow \frac{dy}{dx}=3x^2$

${\left(\frac{dy}{dx}\right)}_{\left(-1, -1\right)}=3$

Equation of tangent at $(–1, –1)$

$(y+1)=3(x+1)$

$y=3x+2 \dots \left(\mathrm{ii}\right)$

Solving $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$

$\Rightarrow x^3=3x+2$

$\Rightarrow x^3-3x+2=0$

$\Rightarrow x=-1,-1,2$

Another point of intersection is $Q(2, 8)$

So, now plotting the diagram we get,

Required area from the above diagram will be,

$={\int }_{-1}^2\left(3x+2-x^3\right)dx$

$=\frac{3}{2}\left(4-1\right)+2\left(2+1\right)-\frac{1}{4}\left(16-1\right)$

$=\frac{27}{4}$

Hence this is the required option.