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Setting up the right-endpoint Riemann sum limit

Consider the function f(x)=1xf(x) = \dfrac{1}{x} on the interval [1,3][1, 3]. We want to express the area under this curve as a limit of right-endpoint Riemann sums.

Divide [1,3][1,3] into nn subintervals of equal width to form a right-endpoint Riemann sum for f(x)=1xf(x) = \dfrac{1}{x}.

  1. Write Δx\Delta x in terms of nn.
  2. Write the right endpoint xix_i of the ii-th subinterval in terms of ii and nn.
  3. Express the exact area under y=1xy = \dfrac{1}{x} on [1,3][1,3] as lim⁡n→∞∑i=1nf(xi) Δx\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n} f(x_i)\,\Delta x, writing f(xi) Δxf(x_i)\,\Delta x explicitly in terms of ii and nn (do not evaluate the limit).

State your final expression for the area as a limit of a sum.

Hint

The interval [1,3][1,3] has length 22, so Δx=2/n\Delta x = 2/n. The ii-th right endpoint is the left endpoint 11 plus ii widths: xi=1+iΔxx_i = 1 + i\Delta x. Substitute this into f(xi)=1/xif(x_i) = 1/x_i and multiply by Δx\Delta x.

Step-by-step solution

Since [1,3][1,3] has length 3−1=23-1=2, splitting into nn equal subintervals gives Δx=2n\Delta x = \dfrac{2}{n}. The right endpoint of the ii-th subinterval is xi=1+iΔx=1+2inx_i = 1 + i\Delta x = 1 + \dfrac{2i}{n}, since we start at x=1x=1 and step forward by Δx\Delta x a total of ii times.

Evaluating ff at this right endpoint gives f(xi)=11+2inf(x_i) = \dfrac{1}{1 + \frac{2i}{n}}, and multiplying by Δx\Delta x gives the term f(xi)Δx=2/n1+2inf(x_i)\Delta x = \dfrac{2/n}{1+\frac{2i}{n}}. Summing over i=1,…,ni=1,\dots,n and taking n→∞n\to\infty gives the exact area: ∫131x dx=lim⁡n→∞∑i=1n2/n1+2in.\int_1^3 \frac{1}{x}\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} \frac{2/n}{1+\frac{2i}{n}}.

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