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Setting up the right-endpoint Riemann sum

Consider the function f(x)=xf(x) = x on the interval [0,5][0, 5]. We want to find the area under the curve using a right-endpoint Riemann sum, then express the exact area as a limit of that sum.

Divide [0,5][0,5] into nn equal subintervals and use the right endpoint of each subinterval as the sample point xi∗x_i^*.

  1. Write Δx\Delta x in terms of nn.
  2. Write xi∗x_i^* in terms of ii and nn.
  3. Write the Riemann sum ∑i=1nf(xi∗) Δx\displaystyle\sum_{i=1}^{n} f(x_i^*)\,\Delta x as an explicit expression in terms of nn only (you may use the fact that ∑i=1ni=n(n+1)2\sum_{i=1}^n i = \frac{n(n+1)}{2}).

What is the simplified closed-form expression for this Riemann sum?

Hint

Start with Δx=5−0n\Delta x = \frac{5-0}{n} and xi∗=0+iΔxx_i^* = 0 + i\Delta x. Substitute f(xi∗)=xi∗f(x_i^*) = x_i^*, factor constants out of the sum, and use the given formula for ∑i=1ni\sum_{i=1}^n i.

Step-by-step solution

With nn equal subintervals on [0,5][0,5], the width is Δx=5n\Delta x = \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}. Since f(x)=xf(x)=x, we have f(xi∗)=5inf(x_i^*) = \frac{5i}{n}.

The Riemann sum is ∑i=1nf(xi∗)Δx=∑i=1n5in⋅5n=25n2∑i=1ni=25n2⋅n(n+1)2=25(n+1)2n\sum_{i=1}^n f(x_i^*)\Delta x = \sum_{i=1}^n \frac{5i}{n}\cdot\frac{5}{n} = \frac{25}{n^2}\sum_{i=1}^n i = \frac{25}{n^2}\cdot\frac{n(n+1)}{2} = \frac{25(n+1)}{2n}.

Taking the limit to get the exact area

Using the closed-form Riemann sum expression from part (a), evaluate lim⁡n→∞∑i=1nf(xi∗)Δx\displaystyle\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x to find ∫05x dx\displaystyle\int_0^5 x\,dx.

Hint

Take the limit of your part (a) expression as n→∞n\to\infty; the term 1/n1/n vanishes.

Step-by-step solution

From part (a), the Riemann sum equals 252(1+1n)\frac{25}{2}\left(1+\frac{1}{n}\right). As n→∞n\to\infty, the term 1n→0\frac{1}{n}\to 0, so the sum approaches 252(1+0)=252\frac{25}{2}(1+0) = \frac{25}{2}.

Therefore ∫05x dx=lim⁡n→∞252(1+1n)=252=12.5\int_0^5 x\,dx = \lim_{n\to\infty}\frac{25}{2}\left(1+\frac1n\right) = \frac{25}{2} = 12.5.

Comparing to the triangle area formula

The region under f(x)=xf(x) = x on [0,5][0,5] is a right triangle with legs along the x-axis and the line y=xy=x.

Setting up the right-endpoint Riemann sum — Comparing to the triangle area formula graph

Compute the area of this triangle using the elementary formula 12base×height\frac{1}{2}\text{base}\times\text{height}, and state how it compares to your answer from part (b).

Hint

Identify the base and height of the triangle from the graph, then compare the resulting number to 252\frac{25}{2}.

Step-by-step solution

The triangle has base 55 (along the x-axis from 00 to 55) and height 55 (the value of f(5)=5f(5)=5), so its area is 12(5)(5)=252=12.5\frac{1}{2}(5)(5) = \frac{25}{2} = 12.5.

This matches exactly the value obtained from the Riemann sum limit in part (b), confirming that the limit-of-Riemann-sums definition of the definite integral gives the same result as the elementary geometric area formula.

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