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Midpoint Riemann sum for cos(x) on [-π/2, π/2] with n = 6

Consider the function f(x)=cos⁡(x)f(x) = \cos(x) on the interval [−π/2,π/2][-\pi/2, \pi/2].

Partition [−π/2,π/2][-\pi/2, \pi/2] into n=6n = 6 subintervals of equal width to approximate ∫−π/2π/2cos⁡(x) dx\int_{-\pi/2}^{\pi/2} \cos(x)\, dx using the midpoint rule. Find the width Δx\Delta x of each subinterval, identify the midpoint of each subinterval, and use these midpoints to compute the midpoint Riemann sum M6=∑i=16f(xˉi) ΔxM_6 = \sum_{i=1}^{6} f(\bar{x}_i)\, \Delta x. What is the value of M6M_6?

Hint

First compute Δx=π/2−(−π/2)6=π6\Delta x = \frac{\pi/2 - (-\pi/2)}{6} = \frac{\pi}{6}. List the six subintervals, find the midpoint of each, evaluate cos⁡\cos at each midpoint, sum the results, and multiply by Δx\Delta x.

Step-by-step solution

The width of each subinterval is Δx=π/2−(−π/2)6=π6\Delta x = \frac{\pi/2 - (-\pi/2)}{6} = \frac{\pi}{6}. The subintervals are [−π/2,−π/3][-\pi/2, -\pi/3], [−π/3,−π/6][-\pi/3, -\pi/6], [−π/6,0][-\pi/6, 0], [0,π/6][0, \pi/6], [π/6,π/3][\pi/6, \pi/3], [π/3,π/2][\pi/3, \pi/2], so the midpoints are −5π12,−π4,−π12,π12,π4,5π12-\frac{5\pi}{12}, -\frac{\pi}{4}, -\frac{\pi}{12}, \frac{\pi}{12}, \frac{\pi}{4}, \frac{5\pi}{12}.

Since cos⁡(x)\cos(x) is even, the cosine values at these symmetric midpoints pair up: cos⁡(5π/12)≈0.2588\cos(5\pi/12) \approx 0.2588, cos⁡(π/4)≈0.7071\cos(\pi/4) \approx 0.7071, cos⁡(π/12)≈0.9659\cos(\pi/12) \approx 0.9659, each appearing twice. Summing gives 2(0.2588+0.7071+0.9659)≈3.86362(0.2588 + 0.7071 + 0.9659) \approx 3.8636.

Multiplying by Δx=π/6≈0.5236\Delta x = \pi/6 \approx 0.5236 gives M6≈3.8636×0.5236≈2.0233M_6 \approx 3.8636 \times 0.5236 \approx 2.0233, which is close to the exact value of the integral, 22.

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