Evaluate `int (2x-5)^2 dx`

Explanation

From the graph ,`f' lt 0 " for " x in (0, 1) cup (5, 6)` where f decreases.
From the graph, ` f' gt 0 " for " x in (1, 5) ` where f increases.
At x = 1, f ' changes sign from '-' to '+' , so it is the point of local minima.
At x = 5, f' changes sign from '+' to '-' , so it is the point of local maxima.
Slope of tangent to curve f' is '+' for ` x in (0, 3) , i.e. f' gt 0`.
Slope of tangent to curve f ' is '-' for ` x in (3, 6), i.e. f' lt 0.`
That is , f'' changes sign at x = 3 with f''(3) = 0, so x = 3 is the point of inflection.