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Setting up the right-endpoint Riemann sum limit

Consider the function f(x)=x2−4x+5f(x) = x^2 - 4x + 5 on the interval [1,3][1, 3].

Divide [1,3][1,3] into nn subintervals of equal width and let xi∗x_i^* denote the right endpoint of the ii-th subinterval. Express the exact area under f(x)=x2−4x+5f(x) = x^2 - 4x + 5 on [1,3][1,3] as a limit of right-endpoint Riemann sums. Write your final expression as a single limit of a sum, fully in terms of nn and ii (i.e., substitute explicit formulas for Δx\Delta x and xi∗x_i^*, and simplify f(xi∗)f(x_i^*) into a polynomial in ii and nn). State the resulting limit expression.

Hint

First find Δx=(b−a)/n\Delta x = (b-a)/n and the right endpoint formula xi∗=a+iΔxx_i^* = a + i\Delta x. Substitute xi∗x_i^* into f(x)f(x), expand and simplify, then write the whole thing as lim⁡n→∞∑i=1nf(xi∗)Δx\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x.

Step-by-step solution

Here a=1a=1, b=3b=3, so Δx=3−1n=2n\Delta x = \frac{3-1}{n} = \frac{2}{n}, and the right endpoint of the ii-th subinterval is xi∗=1+iΔx=1+2inx_i^* = 1 + i\Delta x = 1 + \frac{2i}{n}. The definite integral is by definition ∫13f(x) dx=lim⁡n→∞∑i=1nf(xi∗)Δx\int_1^3 f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x.

Now compute f(xi∗)=(1+2in)2−4(1+2in)+5f(x_i^*) = \left(1+\frac{2i}{n}\right)^2 - 4\left(1+\frac{2i}{n}\right) + 5. Expanding gives 1+4in+4i2n2−4−8in+5=2−4in+4i2n21 + \frac{4i}{n} + \frac{4i^2}{n^2} - 4 - \frac{8i}{n} + 5 = 2 - \frac{4i}{n} + \frac{4i^2}{n^2}. Multiplying by Δx=2n\Delta x = \frac{2}{n} and summing yields the limit expression lim⁡n→∞∑i=1n(2−4in+4i2n2)⋅2n\lim_{n\to\infty}\sum_{i=1}^n \left(2 - \frac{4i}{n} + \frac{4i^2}{n^2}\right)\cdot\frac{2}{n}, which represents the exact area.

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