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

Step-by-step solution

Given y = x 3       . . .   i ⇒ d y d x = 3 x 2 d y d x - 1 ,   - 1 = 3 Equation of tangent at ( – 1 ,   – 1 ) ( y + 1 ) = 3 ( x + 1 ) y = 3 x + 2           …   ( ii ) Solving i and ii ⇒ x 3 = 3 x + 2 ⇒ x 3 - 3 x + 2 = 0 ⇒ 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, = ∫ - 1 2 3 x + 2 - x 3 d x = 3 2 4