# Calculus, integration, Riemann sum help?

Express as a deﬁnite integral and then evaluate the limit of the Riemann sum lim $$\lim_{n\to \infty}\sum_{i=0}^{n-1} (3x_i^2 + 1)\Delta x,$$ where $P$ is the partition with $$x_i = -1 + \frac{3i}{n}$$ for $i = 0, 1, \dots, n$ and $\Delta x \equiv x_i - x_{i-1}$.

I am completely and utterly confused as to how to even start this question. Any help/good links hugely appriciated!

• Please edit the question so it actually includes the question. The link might break at any time, leaving anybody looking here stranded. Also, change the title to something more descriptive. Commented Mar 11, 2013 at 17:42
• Grace: I edited your question. Please check and make sure that I did not alter your question. Also, for future questions, please type up the question. Commented Mar 11, 2013 at 17:47

When $i$ goes from $0$ to $n$, then $-1+\dfrac{3i}{n}$ goes from $-1$ to $2$. There you have the bounds of integration. What you're integrating is $3x^2+1$. So evaluate that integral between those bounds.

• Thanks for the answer @michalhardy but could you explain how you got the -1 and 2 please? Commented Mar 11, 2013 at 17:30
• @GraceBrennan: if $i=0, (-1+\frac {3i}n)=-2$ and if $i=n, (-1+\frac {3i}n)=(-1+\frac {3n}n)=2$ Commented Mar 11, 2013 at 17:32
• @michaelhardy Wow thanks a million! Commented Mar 11, 2013 at 17:34
• @RossMillikan : Your arithmetic seems less than perfect. Commented Mar 11, 2013 at 17:36
• @GraceBrennan : When $i=0$ then $\displaystyle-1+\frac{3i}{n}=-1+0=-1$. When $i=n$ then $\displaystyle-1+\frac{3i}{n}=-1+3=2$. Commented Mar 11, 2013 at 18:24

Let $f$ be a function, and let $[a,b]$ be an interval. Let $n$ be a positive integer, and let $\Delta x=\frac{b-a}{n}$. Let $x_0=a$, $x_1=a+\Delta x$, $x_2=a+2\Delta x$, and so on up to $x_n=a+n\Delta x$. So $x_i=a+i\Delta x$.

So far, a jumble of symbols. You are likely not to ever understand what's going on unless you associate a picture with these symbols.

So draw some nice function $f(x)$, say always positive, and take some interval $[a,b]$. For concreteness, let $a=1$ and $b=4$. Pick a specific $n$, like $n=6$. Then $\Delta x=\frac{3}{6}=\frac{1}{2}$.

So $x_0=1$, $x_1=1.5$, $x_2=2$, $x_3=2.5$, $x_4=3$, $x_5=2.5$, and $x_6=3$. Note that the points divide the interval from $a$ to $b$ into $n$ subintervals. These intervals all have width $\Delta x$.

Now calculate $f(x_0)\Delta x$. This is the area of a certain rectangle. Draw it. Similarly, $f(x_1)\Delta x$ is the area of a certain rectangle. Draw it. Continue up to $f(x_5)\Delta x$. Add up. The sum is called the left Riemann sum associated with the function $f$ and the division of $[1,4]$ into $6$ equal-sized parts.

The left Riemann sum is an approximation to the area under the curve $y=f(x)$, from $x=a$ to $x=b$. Intuitively, if we take $n$ very large, the sum will be a very good approximation to the area, and the limit as $n\to\infty$ of the Riemann sums is the integral $\displaystyle\int_a^b f(x)\,dx$.

Let us apply these ideas to your concrete example. It is basically a matter of pattern recognition. We have $x_0=-1$, $x_1=-1+\frac{3}{n}$, $x_2=-1+\frac{6}{n}$, and so on. These increase by $\frac{3}{n}$, so $\Delta x=\frac{3}{n}$.

We have $x_0=-1$, and $x_n=-1+\frac{3n}{n}=2$. So $a=-1$ and $b=2$.

Our sum is a sum of terms of the shape $(3x_i^2+1)\Delta x$. Comparing with the general pattern $f(x_i)\Delta x$, we see that $f(x)=3x^2+1$.

So for large $n$, the Riemann sum of your problem should be a good approximation to $\displaystyle\int_{-1}^2 (3x^2+1)\,dx$.