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Midpoint Riemann sum vs. exact semicircle area

Consider the function f(x)=9−x2f(x) = \sqrt{9 - x^2} on the interval [−3,3][-3, 3]. Note that the graph of ff is the upper half of the circle x2+y2=9x^2 + y^2 = 9, so the exact area under the curve is the area of a semicircle of radius 33.

Partition [−3,3][-3,3] into n=6n = 6 subintervals of equal width to approximate the area under f(x)=9−x2f(x) = \sqrt{9-x^2}.

  1. Determine Δx\Delta x and the resulting subintervals.
  2. Compute the midpoint sum M6=∑i=16f(mi) ΔxM_6 = \sum_{i=1}^{6} f(m_i)\,\Delta x, where mim_i is the midpoint of the ii-th subinterval.
  3. Compute the exact area of the semicircle using A=12πr2A = \tfrac{1}{2}\pi r^2 with r=3r=3.
Midpoint Riemann sum vs. exact semicircle area graph

Using your midpoint sum, what is the approximate percent difference (in absolute value) between M6M_6 and the exact semicircle area, and does M6M_6 overestimate or underestimate the true area?

Hint

First find Δx=(3−(−3))/6=1\Delta x = (3-(-3))/6 = 1, list the six subintervals, and evaluate ff at each midpoint. Sum these values and multiply by Δx\Delta x to get M6M_6; then compare to A=12π(3)2A = \frac{1}{2}\pi(3)^2.

Step-by-step solution

With Δx=3−(−3)6=1\Delta x = \frac{3-(-3)}{6} = 1, the subintervals are [−3,−2],[−2,−1],[−1,0],[0,1],[1,2],[2,3][-3,-2], [-2,-1], [-1,0], [0,1], [1,2], [2,3], with midpoints −2.5,−1.5,−0.5,0.5,1.5,2.5-2.5, -1.5, -0.5, 0.5, 1.5, 2.5. Evaluating f(x)=9−x2f(x)=\sqrt{9-x^2} at each midpoint gives approximately f(−2.5)≈1.658f(-2.5)\approx1.658, f(−1.5)≈2.598f(-1.5)\approx2.598, f(−0.5)≈2.958f(-0.5)\approx2.958, f(0.5)≈2.958f(0.5)\approx2.958, f(1.5)≈2.598f(1.5)\approx2.598, f(2.5)≈1.658f(2.5)\approx1.658.

Summing these values gives approximately 14.42714.427, and multiplying by Δx=1\Delta x = 1 gives M6≈14.427M_6 \approx 14.427. The exact semicircle area is A=12π(3)2=9π2≈14.137A = \frac{1}{2}\pi(3)^2 = \frac{9\pi}{2} \approx 14.137, so M6M_6 slightly overestimates the true area, with a percent difference of about ∣14.427−14.13714.137∣×100%≈2.05%\left|\frac{14.427-14.137}{14.137}\right|\times100\% \approx 2.05\%.

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