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Ranking Left, Right, and Midpoint sums by shape of the graph

The graph below shows a function ff that is decreasing and concave up on the interval [2,10][2, 10].

Consider the function ff graphed below on [2,10][2, 10], which is decreasing and concave up on this interval. Suppose we approximate the area under ff on [2,10][2,10] using n=4n = 4 subintervals, computing the Left-endpoint sum L4L_4, the Right-endpoint sum R4R_4, and the Midpoint sum M4M_4.

Ranking Left, Right, and Midpoint sums by shape of the graph graph

Using only the facts that ff is decreasing and concave up on [2,10][2,10] (no formula for ff is given), rank L4L_4, R4R_4, and M4M_4 from smallest to largest.

Hint

Think about what decreasing means for comparing left- and right-endpoint sums, and what concave up means for comparing the midpoint sum to the true area (and hence to the other two sums).

Step-by-step solution

Since ff is decreasing on [2,10][2,10], each left endpoint gives a larger function value than the corresponding right endpoint on the same subinterval, so the Left sum overestimates the area and the Right sum underestimates it. This immediately tells us R4<L4R_4 < L_4, since every rectangle used in R4R_4 is shorter than the corresponding rectangle in L4L_4.

Since ff is also concave up, we can apply the property that for a concave up function, the secant line connecting two points on the curve lies below the curve itself. On each subinterval, the midpoint approximation uses the function value at the midpoint, which lies above the secant line connecting the endpoints. Therefore, M4M_4 overestimates the true area. However, M4M_4 overestimates by less than L4L_4 does, because the left-endpoint sum uses the larger endpoint values at the left of each subinterval, while the midpoint uses intermediate values. Combining both observations gives the ranking R4<M4<L4R_4 < M_4 < L_4.

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