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Right-endpoint Riemann sum in sigma notation

Consider the function f(x)=3x+1f(x) = 3x + 1 on the interval [0,5][0,5], divided into nn equal subintervals.

For f(x)=3x+1f(x) = 3x+1 on [0,5][0,5], partition the interval into nn equal subintervals. Write the right-endpoint Riemann sum RnR_n using sigma notation (in terms of nn and ii), then simplify it to a closed-form expression in terms of nn only using the formula ∑i=1ni=n(n+1)2\sum_{i=1}^n i = \frac{n(n+1)}{2}. What is this closed-form expression for RnR_n?

Hint

First find Δx=5−0n\Delta x = \frac{5-0}{n} and the right endpoint xi=0+iΔxx_i = 0 + i\Delta x. Substitute into f(xi)f(x_i), form the sum ∑i=1nf(xi)Δx\sum_{i=1}^n f(x_i)\Delta x, and use linearity of sigma notation plus the sum formula for ∑i\sum i to simplify.

Step-by-step solution

The width of each subinterval is Δx=5−0n=5n\Delta x = \frac{5-0}{n} = \frac{5}{n}, and the right endpoint of the ii-th subinterval is xi=0+iΔx=5inx_i = 0 + i\Delta x = \frac{5i}{n}. Substituting into ff gives f(xi)=3(5in)+1=15in+1f(x_i) = 3\left(\frac{5i}{n}\right) + 1 = \frac{15i}{n} + 1. The right-endpoint sum is therefore Rn=∑i=1nf(xi)Δx=∑i=1n(15in+1)5n.R_n = \sum_{i=1}^n f(x_i)\Delta x = \sum_{i=1}^n \left(\frac{15i}{n}+1\right)\frac{5}{n}.

Expanding and using linearity, Rn=5n∑i=1n(15in+1)=75n2∑i=1ni+5n∑i=1n1.R_n = \frac{5}{n}\sum_{i=1}^n\left(\frac{15i}{n}+1\right) = \frac{75}{n^2}\sum_{i=1}^n i + \frac{5}{n}\sum_{i=1}^n 1. Using ∑i=1ni=n(n+1)2\sum_{i=1}^n i = \frac{n(n+1)}{2} and ∑i=1n1=n\sum_{i=1}^n 1 = n, this becomes Rn=75n2⋅n(n+1)2+5n⋅n=75(n+1)2n+5.R_n = \frac{75}{n^2}\cdot\frac{n(n+1)}{2} + \frac{5}{n}\cdot n = \frac{75(n+1)}{2n} + 5.

Limit of the Riemann sum as n approaches infinity

Using your closed-form expression for RnR_n from part (a), evaluate lim⁡n→∞Rn\displaystyle\lim_{n\to\infty} R_n. What is the value of this limit?

Hint

As n→∞n \to \infty, any term with 1n\frac{1}{n} or 1n2\frac{1}{n^2} vanishes. Simplify RnR_n first, then take the limit term by term.

Step-by-step solution

From part (a), Rn=75(n+1)2n+5=752+752n+5R_n = \frac{75(n+1)}{2n} + 5 = \frac{75}{2} + \frac{75}{2n} + 5. As n→∞n \to \infty, the term 752n\frac{75}{2n} approaches 00.

Taking the limit gives lim⁡n→∞Rn=752+0+5=852.\lim_{n\to\infty} R_n = \frac{75}{2} + 0 + 5 = \frac{85}{2}. This matches the exact area under f(x)=3x+1f(x)=3x+1 on [0,5][0,5], confirming the Riemann sum converges to the definite integral.

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