The area of the region bounded by the curve y = x^3 - 3x + 2 and its tangent line at the point where x = 2 is :

Step-by-step solution

First, we find the equation of the tangent line to the curve y = x^3 - 3x + 2 at x = 2 . When x = 2 , y = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4 . So the point of tangency is (2, 4) . The derivative is dy dx = 3x^2 - 3 . At x = 2 , the slope of the tangent is m = 3(2)^2 - 3 = 12 - 3 = 9 . The equation of the tangent line is y - 4 = 9(x - 2) y = 9x - 14 . Next, we find the points of intersection between the curve and the tangent line by equating them: x^3 - 3x + 2 = 9x - 14 x^3 - 12x + 16 = 0 Since the line is tangent to