22. Riemann Sums, Integrals and the FTC

Recall, in general: \(\displaystyle \int_a^bf(x)\,dx =\lim_{\begin{aligned}&\scriptstyle\quad n\to\infty \\ &\scriptstyle\text{max}\Delta x_i\to0\end{aligned}} \sum_{i=1}^n f(x_i^*)\Delta x_i\) where   \(\Delta x_i=x_i-x_{i-1}\)   and   \(x_i^*\in[x_{i-1},x_i]\). For equal intervals and right endpoints: \(\displaystyle \int_a^bf(x)\,dx =\lim_{n\to\infty} \sum_{i=1}^n f(x_i)\Delta x\)   where   \(\Delta x=\dfrac{b-a}{n}\)   and   \(x_i=a+i\Delta x\).

a3. Limits of Riemann Sums (Optional)

So far, we have approximated integrals using equal intervals with right endpoints, or left endpoints or midpoints. We now discuss how to actually compute the integrals by taking the limits. We will do this by example and restrict to polynomial integrands and use equal intervals and right endpoints.

We first recall some basic summation formulas from high school algebra or pre-calculus.

\[\begin{aligned} \sum_{i=1}^n 1 &=n &\sum_{i=1}^n i &=\dfrac{n(n+1)}{2} \\ \sum_{i=1}^n i^2 &=\dfrac{n(n+1)(2n+1)}{6} \qquad &\sum_{i=1}^n i^3 &=\dfrac{n^2(n+1)^2}{4} \end{aligned}\] These are proved using Mathematical Induction.

We split the interval \([1,3]\) into \(n\) equal parts of width: \[ \Delta x=\dfrac{3-1}{n}=\dfrac{2}{n} \] Then the right endpoints are: \[ x_i=1+i\Delta x=1+\dfrac{2}{n}i \] Since the integrand is the function \(f(x)=x^3-6x\), its value at a right endpoint is: \[\begin{aligned} f(x_i)&=(x_i)^3=\left(1+\dfrac{2}{n}i\right)^3-6\left(1+\dfrac{2}{n}i\right) \\ &=1+3\dfrac{2}{n}i+3\dfrac{2^2}{n^2}i^2+\dfrac{2^3}{n^3}i^3-6-\dfrac{12}{n}i \\ &=-5-\dfrac{6}{n}i+\dfrac{12}{n^2}i^2+\dfrac{8}{n^3}i^3 \end{aligned}\] We set up the Riemann sum, expand each term, write it as a sum of summations and factor out the coefficients: \[\begin{aligned} \sum_{i=1}^n &f(x_i)\Delta x =\sum_{i=1}^n \left[\left(1+\dfrac{2}{n}i\right)^3-6\left(1+\dfrac{2}{n}i\right)\right]\dfrac{2}{n} \\ &=\sum_{i=1}^n \left(-5-\dfrac{6}{n}i+\dfrac{12}{n^2}i^2+\dfrac{8}{n^3}i^3\right) \dfrac{2}{n} \\ &=\sum_{i=1}^n \left(-\dfrac{10}{n}-\dfrac{12}{n^2}i +\dfrac{24}{n^3}i^2+\dfrac{16}{n^4}i^3\right) \\ &=\sum_{i=1}^n \dfrac{-10}{n}+\sum_{i=1}^n \dfrac{-12}{n^2}i +\sum_{i=1}^n \dfrac{24}{n^3}i^2+\sum_{i=1}^n \dfrac{16}{n^4}i^3 \\ &=-\,\dfrac{10}{n}\sum_{i=1}^n 1-\dfrac{12}{n^2}\sum_{i=1}^n i +\dfrac{24}{n^3}\sum_{i=1}^n i^2+\dfrac{16}{n^4}\sum_{i=1}^n i^3 \end{aligned}\] We now use the basic summations and simplify: \[\begin{aligned} \sum_{i=1}^n &f(x_i)\Delta x \\ &=-\,\dfrac{10}{n}\cdot n-\dfrac{12}{n^2}\cdot\dfrac{n(n+1)}{2} +\dfrac{24}{n^3}\cdot\dfrac{n(n+1)(2n+1)}{6} +\dfrac{16}{n^4}\cdot\dfrac{n^2(n+1)^2}{4} \\ &=-10-6\dfrac{(n+1)}{n} +4\dfrac{(n+1)(2n+1)}{n^2} +4\dfrac{(n+1)^2}{n^2} \\ \end{aligned}\] Finally, we compute the integral as the limit of this Riemann sum: \[\begin{aligned} \int_1^3 &x^3-6x\,dx =\lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x \\ &=\lim_{n\to\infty}\left(-10-6\dfrac{(n+1)}{n} +4\dfrac{(n+1)(2n+1)}{n^2}+4\dfrac{(n+1)^2}{n^2}\right) \\ &=-10-6+8+4=-4 \end{aligned}\]