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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!

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    $\begingroup$ 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. $\endgroup$
    – vonbrand
    Commented Mar 11, 2013 at 17:42
  • $\begingroup$ 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. $\endgroup$
    – Thomas
    Commented Mar 11, 2013 at 17:47

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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.

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

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