Figure N6–8 is helpful in showing how the area of a midpoint rectangle
compares with that of a trapezoid and with the true area. Our graph here is
concave down. If M is the midpoint of AB, then the midpoint rectangle is
AM 1 M 2 B. We’ve drawn T 1 T 2 tangent to the curve at T (where the midpoint
ordinate intersects the curve). Since the shaded triangles have equal areas, we
see that area AM 1 M 2 B = area AT 1 T 2 B.† But area AT 1 T 2 B clearly exceeds the true
area, as does the area of the midpoint rectangle. This fact justifies the right half
of the inequality below; Figure N6–7 verifies the left half.
Figure N6–8Generalizing to n subintervals, we conclude:
If the graph of f is concave down, then
.If the graph of f is concave up, then† Note that the trapezoid AT 1 T 2 B is different from the trapezoids in Figure N6–7, which are like the ones
we use in applying the trapezoid rule.Example 27 __
Write an inequality including L(n), R(n), M(n), T(n), and for the graph of
f shown in Figure N6–9.