We are estimating the area under the graph of f(x)=1x2+2 f(x) = \frac{1}{x^2 + 2} f(x)=x2+21 over the interval [3,6] [3, 6] [3,6] using n=5 n = 5 n=5 rectangles.
The width Δx \Delta x Δx of each rectangle is calculated as: Δx=b−an=6−35=0.6 \Delta x = \frac{b - a}{n} = \frac{6 - 3}{5} = 0.6 Δx=nb−a=56−3=0.6
Using the right endpoints, we evaluate the function at the points: x1=3.6,x2=4.2,x3=4.8,x4=5.4,x5=6.0 x_1 = 3.6, \quad x_2 = 4.2, \quad x_3 = 4.8, \quad x_4 = 5.4, \quad x_5 = 6.0 x1=3.6,x2=4.2,x3=4.8,x4=5.4,x5=6.0 The Riemann sum Rn R_n Rn is given by: Rn=∑i=1nf(xi)Δx R_n = \sum_{i=1}^{n} f(x_i) \Delta x Rn=i=1∑nf(xi)Δx Calculating Rn R_n Rn: Rn≈0.1297 R_n \approx 0.1297 Rn≈0.1297
Using the left endpoints, we evaluate the function at the points: x0=3.0,x1=3.6,x2=4.2,x3=4.8,x4=5.4 x_0 = 3.0, \quad x_1 = 3.6, \quad x_2 = 4.2, \quad x_3 = 4.8, \quad x_4 = 5.4 x0=3.0,x1=3.6,x2=4.2,x3=4.8,x4=5.4 The Riemann sum Ln L_n Ln is given by: Ln=∑i=0n−1f(xi)Δx L_n = \sum_{i=0}^{n-1} f(x_i) \Delta x Ln=i=0∑n−1f(xi)Δx Calculating Ln L_n Ln: Ln≈0.1684 L_n \approx 0.1684 Ln≈0.1684
The area under the graph using right endpoints is approximately Rn≈0.1297 R_n \approx 0.1297 Rn≈0.1297 and using left endpoints is approximately Ln≈0.1684 L_n \approx 0.1684 Ln≈0.1684.
Thus, the final answers are: Rn≈0.1297 \boxed{R_n \approx 0.1297} Rn≈0.1297 Ln≈0.1684 \boxed{L_n \approx 0.1684} Ln≈0.1684
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