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Midpoint Riemann sum with n = 4

Consider the function f(x)=1x2+1f(x) = \dfrac{1}{x^2+1} on the interval [0,2][0,2].

Partition [0,2][0,2] into n=4n = 4 subintervals of equal width, and use the midpoint of each subinterval to approximate the area under f(x)=1x2+1f(x) = \dfrac{1}{x^2+1} on [0,2][0,2]. Compute the midpoint sum M4M_4.

Hint

First find Δx=(2−0)/4\Delta x = (2-0)/4, list the four subintervals, and find the midpoint of each one. Then evaluate ff at each midpoint and multiply the sum of those values by Δx\Delta x.

Step-by-step solution

The width of each subinterval is Δx=2−04=0.5\Delta x = \frac{2-0}{4} = 0.5, giving subintervals [0,0.5],[0.5,1],[1,1.5],[1.5,2][0,0.5], [0.5,1], [1,1.5], [1.5,2]. The midpoints of these subintervals are x=0.25,0.75,1.25,1.75x = 0.25, 0.75, 1.25, 1.75.

Evaluating ff at each midpoint: f(0.25)=11.0625≈0.9412f(0.25) = \frac{1}{1.0625} \approx 0.9412, f(0.75)=11.5625≈0.64f(0.75) = \frac{1}{1.5625} \approx 0.64, f(1.25)=12.5625≈0.3902f(1.25) = \frac{1}{2.5625} \approx 0.3902, f(1.75)=14.0625≈0.2462f(1.75) = \frac{1}{4.0625} \approx 0.2462.

The midpoint sum is M4=Δx[f(0.25)+f(0.75)+f(1.25)+f(1.75)]≈0.5(0.9412+0.64+0.3902+0.2462)≈0.5(2.2176)≈1.1088M_4 = \Delta x \left[f(0.25)+f(0.75)+f(1.25)+f(1.75)\right] \approx 0.5(0.9412+0.64+0.3902+0.2462) \approx 0.5(2.2176) \approx 1.1088.

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