Answer to Question #260494 in Calculus for Roots

Question

Consider the function f(x)=(x−2)^3

(a) Estimate the area between the curve y=f(x), the x-axis, and the lines x=2 and x=6, using a Riemann sum with four rectangles (use the right endpoint rule).

(b) Calculate the precise area by taking the limit of a Riemann sum.

Answer 1

$\newcommand{\disp}{\displaystyle} \disp f(x )= (x-2)^3\\ \text{Our change in x($\Delta$) is given by}\\ \Delta x= \frac{6-2}{4}=1\\ \text{The Riemann sum which we denote by $S_n$ is}\\ S_n = \Delta x[f(2+1) +f(2+2) + f(2+3)+f(2+4)]\\ S_n = 1[f(3) +f(4) +f(5) +f(6)]\\ = (3-2)^3+(4-2)^3+(5-2)^3+(6-2)^3=100\\ \text{Suppose [2,6] is divided into n intervals with s the width of each}\\ \text{interval $\frac{6-2}{n} = \frac{4}{n}$}\\ \text{The area of each sum is $\frac{4}{n}(x_i-2)^3$, i = 1, 2, ..., n and}\\ x_i = \frac{4i}{n}\\ \text{Since we are using the right Riemann sum, we have}\\ S = \sum^n_{i=1}\frac{4}{n}(\frac{4i}{n}-2)^3\\ = \frac{1}{n^4}(\sum^n_{i=1} 64i^3-96i^2n+81n^2-8n^3)\\ = \frac{1}{n^4}(64(\frac{n(n+1)}{2})^2-96n(\frac{n(n+1)(2n+1)}{6})+8n^2(\frac{n(n+1)}{2})-8n^4)\\ \text{Taking the limit of the expression above, we have that}\\ \lim_{n \to \infty} \frac{1}{n^4}(64(\frac{n(n+1)}{2})^2-96n(\frac{n(n+1)(2n+1)}{6})+8n^2(\frac{n(n+1)}{2})-8n^4)=12$