I know this is a basic integration question, but I am just beginning to learn the subject and don't fully understand it yet - which is why I don't know the answer to this question.
The question is:
For $f(x) = x^{2}$, divide the interval [0,2] into $n$ equally-wide subintervals and evaluate the lower sum and the limit of the lower sum as $ n \rightarrow \infty$.

I know that if $ n \rightarrow \infty$ then I am taking the limit of the Riemann Sum - or using the integral. But, how would I solve this problem step by step?


Using a Riemann Sum (With Right Endpoints):

$$\lim_{n \to \infty}(\frac{b-a}{n}\sum_{i=1}^nf(a +i(\frac{b-a}{n})))$$ $$\lim_{n \to \infty}(\frac{2}{n}\sum_{i=1}^n(i(\frac{2}{n}))^2)$$ $$\lim_{n \to \infty}(\frac{2}{n}\sum_{i=1}^n(\frac{4i^2}{n^2}))$$ Factor out the constants $$\lim_{n \to \infty}(\frac{8}{n^3}\sum_{i=1}^n(i^2))$$ Using the summation property of $i^2$ $$\lim_{n \to \infty}(\frac{8}{n^3}(\frac{n(n+1)(2n+1)}{6}))$$ Distribute the n $$\lim_{n \to \infty}(\frac{8}{n^3}(\frac{(2n^3+3n^2+n)}{6}))$$ $$\lim_{n \to \infty}(\frac{8(2n^3+3n^2+n)}{6n^3})$$ Because this limit tends to infinity, we look at the highest-degree n's. $$\lim_{n \to \infty}\frac{16n^3}{6n^3}$$ $$\lim_{n \to \infty}\frac{8}{3}$$ $$\frac{8}{3}$$

Summation Properties $$\sum_{i=1}^n cf = c*\sum_{i=1}^n f$$ Where c is any constant $$\sum_{i=1}^n c = cn $$ $$\sum_{i=1}^n f+g =\sum_{i=1}^nf + \sum_{i=1}^ng $$ $$\sum_{i=1}^n i = \frac{n(n+1)}{2} $$ $$\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$$ $$\sum_{i=1}^n i^3 = (\frac{n(n+1)}{2})^2$$

  • $\begingroup$ What is the summation property of $i^{2}$? $\endgroup$ – Burt Jul 17 '19 at 3:29
  • $\begingroup$ $\frac{n(n+1)(2n+1)}{6}$ $\endgroup$ – N. Bar Jul 17 '19 at 3:30
  • $\begingroup$ OP asked for lower sum $\endgroup$ – J. W. Tanner Jul 17 '19 at 4:40
  • $\begingroup$ @N.Bar - you offered to list some summation properties. If you don't mind, that would really help me out. $\endgroup$ – Burt Jul 17 '19 at 14:10


$$\sum_{i=0}^{n-1}(i\Delta x)^2\Delta x= (\Delta x)^3\sum_{i=0}^{n-1}i^2=\left(\dfrac2n\right)^3\dfrac{(n-1)n(2n-1)}{6}$$

  • $\begingroup$ Why is it $(i \Delta x)^{2}$ and not just $x^{2}$ $\endgroup$ – Burt Jul 17 '19 at 3:17
  • $\begingroup$ Think of $i\Delta x$ as $x$. As $i$ ranges from $0$ to $n$, $i\Delta x$ ranges from $0$ to $2$, with $\Delta x=\dfrac 2n$ $\endgroup$ – J. W. Tanner Jul 17 '19 at 3:21
  • $\begingroup$ Note: $\dfrac{(n-1)n(2n-1)}{n^3}=\dfrac{2n^3-3n^2+n}{n^3}=2-\dfrac3n+\dfrac1{n^2}$ $\endgroup$ – J. W. Tanner Jul 17 '19 at 3:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.