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Right-endpoint Riemann sum in sigma notation

Consider the function f(x)=x2f(x) = x^2 on the interval [0,3][0,3], partitioned into n=6n = 6 equal subintervals.

The interval [0,3][0,3] is divided into n=6n = 6 equal subintervals to form a right-endpoint Riemann sum for f(x)=x2f(x) = x^2. Determine Δx\Delta x and the right endpoint xix_i in terms of ii, then write the Riemann sum ∑i=16f(xi)Δx\sum_{i=1}^{6} f(x_i)\Delta x using sigma notation with f(xi)f(x_i) expressed explicitly in terms of ii (do not evaluate the sum numerically).

Hint

First find Δx=(b−a)/n\Delta x = (b-a)/n, then write the right endpoint xi=a+iΔxx_i = a + i\Delta x. Substitute this into f(xi)=xi2f(x_i) = x_i^2 and multiply by Δx\Delta x before summing over ii from 1 to 6.

Step-by-step solution

With a=0a=0, b=3b=3, and n=6n=6, the width of each subinterval is Δx=3−06=12\Delta x = \frac{3-0}{6} = \frac{1}{2}. The right endpoint of the ii-th subinterval is xi=a+iΔx=i2x_i = a + i\Delta x = \frac{i}{2}, for i=1,2,…,6i = 1, 2, \dots, 6.

Substituting into f(x)=x2f(x) = x^2 gives f(xi)=(i2)2=i24f(x_i) = \left(\frac{i}{2}\right)^2 = \frac{i^2}{4}. Multiplying by Δx=12\Delta x = \frac{1}{2}, the Riemann sum becomes ∑i=16i24⋅12=∑i=16i28.\sum_{i=1}^{6} \frac{i^2}{4}\cdot\frac{1}{2} = \sum_{i=1}^{6} \frac{i^2}{8}.

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