Calculus estimate please help

Question

Given f(x) > 0 with f ′ (x) < 0, and f ″(x) < 0 for all x in the interval [0, 1] with f(0) = 1 and f(1) = 0.3, the left, right, trapezoidal, and midpoint rule approximations were used to estimate . The estimates were 0.7915, 0.8405, 0.8410, 0.8421 and 0.8895, and the same number of subintervals were used in each case. Match the rule to its estimate.

1. Right endpoint	0.8895
2.Left endpoint	0.7915
3.Midpoint	0.8421
4.actual area	0.8405
5.trapezoidal	0.8410

Answer 1

As f'(x) < 0 throughout the interval, the max of these values will be the left endpoint, and the min will be the left endpoint. As f'' < 0, the curve and midpoint is above the trapezoid. Then, consider the trapezoid formed using the midpoint with the slope equaling the slope at the midpoint; the area of this is the same as the area with the midpoint rule. As f'' < 0, the actual curve will be below the trapezoid, so the area will be less. Thus, we now have our order.

From max to min, it is left, midpoint, actual, trapezoid, right.

Thus, the answer is left .8895; midpoint .8421; actual .8410; trapezoid .8405; right .7915

Reordering to match the problem, we have

right .7915

left .8895

midpoint .8421

actual .8410

trapezoid (1 + .3)/2 = .65

The order

trapezoid .8405

As an example to show the order is reasonable, consider f(x) = 1 - .7x^2 and one interval; we have the right curvature and proper values of f(0) and f(1), with f' < 0 for x > 0.

right .3

left 1

midpoint 1 - .7(.5)^2 = .825

actual using I[a,b] for the integral from a to b and E[a, b] the evaluation from a to b

I[0,1] 1 - .7x^2 dx = x - 7/30x^3 E[0, 1] = 23/30 = .76667

trapezoid (1 + .3)/2 = .65

Order from larges to smallest is left, midpoint, actual, trapezoid, and right, just as we said.