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Setting up the right-endpoint Riemann sum limit for the area under ln(x)

Consider the function f(x)=ln⁡(x)f(x) = \ln(x) on the interval [1,e][1, e].

The region under f(x)=ln⁡(x)f(x) = \ln(x) on [1,e][1, e] is divided into nn subintervals of equal width, and the right endpoint of each subinterval is used as the sample point.

area=∫1eln⁡(x) dx\text{area} = \int_1^e \ln(x)\,dx

Write this area exactly as a limit of right-endpoint Riemann sums. Your answer should express Δx\Delta x and xi∗x_i^* explicitly in terms of nn and ii, and give the full expression lim⁡n→∞∑i=1nf(xi∗) Δx\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\,\Delta x with f(xi∗)f(x_i^*) written out explicitly (do not evaluate the limit).

Hint

First find Δx=b−an\Delta x = \frac{b-a}{n} for a=1a=1, b=eb=e. Then the right endpoint of the ii-th subinterval is xi∗=a+iΔxx_i^* = a + i\Delta x. Substitute both into f(xi∗)=ln⁡(xi∗)f(x_i^*) = \ln(x_i^*) and assemble the sum inside the limit.

Step-by-step solution

Since the interval is [1,e][1,e] split into nn equal pieces, the width of each subinterval is Δx=e−1n\Delta x = \dfrac{e-1}{n}. The right endpoint of the ii-th subinterval, starting from a=1a=1, is xi∗=1+iΔx=1+i(e−1)nx_i^* = 1 + i\Delta x = 1 + \dfrac{i(e-1)}{n}.

Substituting this sample point into f(x)=ln⁡(x)f(x) = \ln(x) gives f(xi∗)=ln⁡(1+i(e−1)n)f(x_i^*) = \ln\left(1 + \dfrac{i(e-1)}{n}\right). The definite integral is then the limit of the Riemann sum: ∫1eln⁡(x) dx=lim⁡n→∞∑i=1nln⁡(1+i(e−1)n)⋅e−1n.\int_1^e \ln(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^n \ln\left(1+\frac{i(e-1)}{n}\right)\cdot\frac{e-1}{n}.

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