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Set up the right-endpoint Riemann sum and limit for area under $e^x$

Consider the function f(x)=exf(x) = e^x on the interval [0,1][0,1]. We want to express the area under the curve using a Riemann sum with nn equal subintervals and right endpoints as sample points.

Divide [0,1][0,1] into nn equal subintervals, and let xi∗x_i^* be the right endpoint of the ii-th subinterval. Write Δx\Delta x in terms of nn, write xi∗x_i^* in terms of ii and nn, and then express the area under f(x)=exf(x) = e^x on [0,1][0,1] as a limit of Riemann sums: ∫01ex dx=lim⁡n→∞∑i=1nf(xi∗) Δx.\int_0^1 e^x\,dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*)\,\Delta x. Give the explicit formula for Δx\Delta x, the explicit formula for xi∗x_i^*, and the full limit expression with f(xi∗)Δxf(x_i^*)\Delta x written out in terms of ii and nn (you do not need to evaluate the limit).

Hint

Recall that for nn equal subintervals of [a,b][a,b], Δx=b−an\Delta x = \frac{b-a}{n}, and the right endpoint of the ii-th subinterval is xi∗=a+iΔxx_i^* = a + i\Delta x. Substitute these into f(x)=exf(x) = e^x but do not attempt to evaluate the sum or limit.

Step-by-step solution

Since [0,1][0,1] is divided into nn equal subintervals, each has width Δx=1−0n=1n\Delta x = \frac{1-0}{n} = \frac{1}{n}. The right endpoint of the ii-th subinterval, starting from a=0a=0, is xi∗=0+iΔx=inx_i^* = 0 + i\Delta x = \frac{i}{n}.

Substituting into f(x)=exf(x) = e^x gives f(xi∗)=ei/nf(x_i^*) = e^{i/n}, so each term of the Riemann sum is f(xi∗)Δx=ei/n⋅1nf(x_i^*)\Delta x = e^{i/n}\cdot \frac{1}{n}. Therefore the area is ∫01ex dx=lim⁡n→∞∑i=1nei/n⋅1n.\int_0^1 e^x\,dx = \lim_{n\to\infty}\sum_{i=1}^n e^{i/n}\cdot\frac{1}{n}. This expresses the exact area under the curve as the limit of the sum of areas of thin rectangles whose heights are given by the right endpoints.

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