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Right Riemann sum with n = 6 subintervals

Consider the function f(x)=9−x2f(x) = 9 - x^2 on the interval [0,3][0, 3].

Partition [0,3][0,3] into n=6n = 6 equal subintervals and use the right Riemann sum to approximate ∫03(9−x2) dx\displaystyle\int_0^3 (9 - x^2)\, dx.

First determine Δx\Delta x and the right endpoints x1,x2,…,x6x_1, x_2, \ldots, x_6, then compute R6=∑i=16f(xi) Δx.R_6 = \sum_{i=1}^{6} f(x_i)\,\Delta x. What is the value of R6R_6?

Hint

Since [0,3][0,3] is split into 6 equal pieces, Δx=0.5\Delta x = 0.5. The right endpoints are xi=0.5ix_i = 0.5i for i=1,…,6i = 1, \ldots, 6. Evaluate ff at each of these six points, add them up, then multiply the total by Δx\Delta x.

Step-by-step solution

With n=6n=6 subintervals on [0,3][0,3], the width is Δx=3−06=0.5\Delta x = \frac{3-0}{6} = 0.5. The right endpoints are x1=0.5, x2=1, x3=1.5, x4=2, x5=2.5, x6=3x_1 = 0.5,\ x_2 = 1,\ x_3 = 1.5,\ x_4 = 2,\ x_5 = 2.5,\ x_6 = 3. The right Riemann sum is R6=Δx[f(x1)+f(x2)+⋯+f(x6)]R_6 = \Delta x \left[ f(x_1) + f(x_2) + \cdots + f(x_6)\right].

Evaluating f(x)=9−x2f(x) = 9 - x^2 at each right endpoint gives f(0.5)=8.75f(0.5)=8.75, f(1)=8f(1)=8, f(1.5)=6.75f(1.5)=6.75, f(2)=5f(2)=5, f(2.5)=2.75f(2.5)=2.75, f(3)=0f(3)=0. Summing these values gives 8.75+8+6.75+5+2.75+0=31.258.75+8+6.75+5+2.75+0 = 31.25. Multiplying by Δx=0.5\Delta x = 0.5 gives R6=0.5×31.25=15.625R_6 = 0.5 \times 31.25 = 15.625.

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