Why is the approximate integral by the midpoint rule the area of a trapazoid and not a rectangle?

If you pivot the top side of the rectangle that represents the midpoint approximation about its midpoint, the area under the line doesn't change. If the function you are integrating is differentiable at the midpoint, you can pivot that line so that it is the tangent line to the graph at the midpoint. If f" > 0 on the interval of integration, the resulting line is strictly below the graph of the function. What does this say about approximations of the integral by the midpoint rule?

Created by Mathematica  (April 4, 2004)