Given f(x) > 0 with f ′ (x) > 0, and f ″ (x) > 0 for all x in the interval [0, 3] with f(0) = 0.1 and f(3) = 1, the left, right, trapezoidal, and midpoint rule approximations...

Question

Given f(x) > 0 with f ′ (x) > 0, and f ″ (x) > 0 for all x in the interval [0, 3] with f(0) = 0.1 and f(3) = 1, the left, right, trapezoidal, and midpoint rule approximations were used to estimate . The estimates were 0.8067, 0.9635, 1.0514, 1.0753 and 1.3439, and the same number of subintervals were used in each case. Match the rule to its estimate.

	1.	trapezoidal
	2.	left endpoint
	3.	actual area
	4.	right end point
	5.	midpoint
a.	0.8067
b.	1.3439
c.	0.9635
d.	1.0753
e.	1.0514

Answer 1

As f'(x) > 0 , we immediately know that the left is the least and the right is the most, with the trapezoidal value in the middle. f''(x) > 0 tells us that the value at the midpoint is under the line between the endpoints and the curve is under the line, so actual area and midpoint values are less than the trapezoid and above the left endpoint because f' > 0. Then, the area from the midpoints is equivalent to the area of trapezoids constructed using the tangent to the midpoint. As f'' > 0, the curve will be above this tangent, so the actual area is above the midpoint estimate.

Thus, we have 2. left < 5. midpoint < 3. actual < 1. trapezoid < 4. right

As a < c < e < d < b, we have

2 a

5 c

3 e

1 d

4 b

If you wish to reorganize,

1 d

2 a

3 e

5 c

4 b