The area of the region enclosed by the curve y = x ^ 3 and its...

Given $y = {x}^{3} . . . \left(i\right)</p><p>$⇒ $\dfrac{dy}{dx} =$3{x}^{2}$</p><p>${\left(\dfrac{dy}{dx}\right)}_{\left(- 1, - 1\right)} = 3$</p><p>Equation of tangent at$( – 1, – 1 )$</p><p>$( y + 1 ) = 3 ( x + 1 )$</p><p>$y = 3x + 2 \ldots ( ii )$</p><p>Solving \left(i\right)$ and \left(ii\right)$</p><p>$⇒ {x}^{3} = 3x + 2$</p><p>$⇒ {x}^{3} - 3x + 2 = 0$</p><p>$⇒ x = - 1, - 1, 2$</p><p>Another point of intersection is$Q ( 2, 8 )$</p><p>So, now plotting the diagram we get,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/9c83caf9-e706-4ce9-9047-ac9a5f324688.png" style="width: 314px; height: 237px;" /></p><p>Required area from the above diagram will be,</p><p>$= {∫}_{- 1}^{2}\left(3x + 2 - {x}^{3}\right)dx$</p><p>$= \dfrac{3}{2}\left(4 - 1\right)$+ 2\left(2 + 1\right)$ - $\dfrac{1}{4}\left(16 - 1\right)</p><p>=\dfrac{27}{4}$

Hence this is the required option.