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I am not sure how to do this problem now since the book gives no examples of what to do for two different types of variables.

I have the integral $$\int_{-2}^0(x^2+x)\,dx.$$

I know that $\Delta x = \frac{2}{n}$.

So now I have $\frac{2}{n}\left(-2 +\frac{2k}{n}\right)$

Then I put that into the equation and I get $$\frac{2}{n}\left(-2+\frac{2k}{n}\right)^2 + \left(-2+\frac{2k}{n}\right).$$

Here is where I really get lost, I do not know how to work with a square and a not square, I have factored it many ways and it didn't really help.

2 Answers 2

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In that last line you really want

$$\frac{2}{n}\left(\left(-2+\frac{2k}{n}\right)^2 + \left(-2+\frac{2k}{n}\right)\right)\;,\tag{1}$$

since the factor of $\dfrac2n$ applies to both terms. In fact, you really want a sum,

$$\sum_{k=1}^n\frac{2}{n}\left(\left(-2+\frac{2k}{n}\right)^2 + \left(-2+\frac{2k}{n}\right)\right)\;,$$

but let’s simplify $(1)$ a bit first.

$$\begin{align*} \left(-2+\frac{2k}{n}\right)^2 + \left(-2+\frac{2k}{n}\right)&=\left(4-\frac{8k}n+\frac{4k^2}{n^2}\right)+\left(-2+\frac{2k}n\right)\\ &=2-\frac{6k}n+\frac{4k^2}{n^2}\;, \end{align*}$$

so $(1)$ is

$$\frac2n\left(2-\frac{6k}n+\frac{4k^2}{n^2}\right)\;,$$ and we want to evaluate

$$\sum_{k=1}^n\frac2n\left(2-\frac{6k}n+\frac{4k^2}{n^2}\right)\;.$$

Now $\dfrac2n$ doesn’t depend on $k$, so you can factor it out of the summation to get $$R_n=\frac2n\sum_{k=1}^n\left(2-\frac{6k}n+\frac{4k^2}{n^2}\right)\;.\tag{2}$$ The trick now is to break up the summation into three simpler ones,

$$\sum_{k=1}^n\left(2-\frac{6k}n+\frac{4k^2}{n^2}\right)=\sum_{k=1}^n2-\sum_{k=1}^n\frac{6k}n+\sum_{k=1}^n\frac{4k^k}{n^2}\;,\tag{3}$$

and evaluate each of those separately. The first one is just $2+2+\ldots+2$, with $n$ terms, so it’s $2n$. The other two you can simplify as you’ve done before to $$\frac6n\sum_{k=1}^nk\quad\text{and}\quad\frac4{n^2}\sum_{k=1}^nk^2\;.$$ Now you want to evaluate these, substitute back into the righthand side of $(3)$, and plug that back into $(2)$ to get $R_n$. Then see if you can calculate $\lim\limits_{n\to\infty}R_n$.

Added: First,

$$\frac6n\sum_{k=1}^nk=\frac6n\cdot\frac{n(n+1)}2=3(n+1)\;,$$

and

$$\frac4{n^2}\sum_{k=1}^nk^2=\frac4{n^2}\cdot\frac{n(n+1)(2n+1)}6=\frac{2(n+1)(2n+1)}{3n}\;,$$

so $(3)$ is $$2n-3(n+1)+\frac{2(n+1)(2n+1)}{3n}=-n-3+\frac{4n^2+6n+2}{3n}\;,$$ and $$\begin{align*}R_n&=\frac2n\left(-n-3+\frac{4n^2+6n+2}{3n}\right)\\ &=-2-\frac6n+\frac{8n^2+12n+4}{3n^2}\\ &=-2-\frac6n+\frac83+\frac4n+\frac4{3n}\;. \end{align*}$$

Now just take the limit as $n\to\infty$.

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    I am getting 4 + 8/3 which I know is wrong. I will type out what I did. Not sure how to really navigate these problems. Seems like there are hundreds of ways to go about these and all the ones I pick are wrong.2012-04-14
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    You probably did at least part of it right to get the $\frac83$.2012-04-14
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    I tried it again and got 7n + 5 which I am pretty sure I need to get rid of ns.2012-04-14
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You seem to be trying to compute a Riemann sum for the integral in question. It's hard to tell because you don't really give enough information (and you complain about the book being unclear...). For example, you never say if you are trying to compute a left hand sum or a right hand sum or what.

Your function is $f(x) = x^2+x$; your interval is $[-2,0]$, You are dividing it into $n$ equal parts, so that the partition will be $$x_0 = -2,\ x_1 = -2+\frac{2}{n},\ x_2 = -2+\frac{4}{n},\ldots,\ x_n = -2+\frac{2n}{n}=0.$$ Then you want to compute either $$\text{LEFT}(n) = \sum_{i=0}^{n-1} f(x_i)\Delta x$$ or $$\text{RIGHT}(n) = \sum_{i=1}^{n}f(x_i)\Delta x.$$

Your formula above is incorrect, since your $\frac{2}{n}$ is only multiplying $x^2$, instead of the entire value $f(x)$.

One thing you can notice is that $\Delta x$ is the same for all $i$, so you can factor it out of your summation. Let me do the Left Hand Sum: $$\begin{align*} \mathrm{LEFT}(n) &= \sum_{i=0}^{n-1}f(x_i)\Delta x\\ &= \sum_{i=0}^{n-1}\left(x_i^2 + x_i\right)\frac{2}{n}\\ &= \frac{2}{n}\sum_{i=0}^{n-1}(x_i^2+x_i)\\ &= \frac{2}{n}\sum_{i=0}^{n-1}\left( \left(-2+\frac{2i}{n}\right)^2 + \left(-2 + \frac{2i}{n}\right)\right)\\ &= \frac{2}{n}\sum_{i=0}^{n-1}\left( (-2)^2 +2(-2)\frac{2i}{n} + \frac{4i^2}{n^2} -2 + \frac{2i}{n}\right)\\ &= \frac{2}{n}\sum_{i=0}^{n-1}\left( 2 - \frac{8i}{n} + \frac{4i^2}{n^2} + \frac{2i}{n}\right)\\ &= \frac{2}{n}\sum_{i=0}^{n-1}\left( 2 - \frac{6i}{n} + \frac{4i^2}{n^2}\right)\\ &= \frac{2}{n}\sum_{i=0}^{n-1}\frac{2n^2 - 6in + 4i^2}{n^2}\\ &= \frac{2}{n^3}\sum_{i=0}^{n-1}\left(2n^2 - 6in + 4i^2\right)\\ &= \frac{2}{n^3}\left( \sum_{i=0}^{n-1}2n^2 - \sum_{i=0}^{n-1}6in + \sum_{i=0}^{n-1}4i^2\right)\\ &= \frac{2}{n^3}\left( 2n^2\sum_{i=0}^{n-1}1 - 6n\sum_{i=0}^{n-1}i + 4\sum_{i=0}^{n-1}i^2\right). \end{align*}$$ Now, the first sum is $$\underbrace{1+1+\cdots+1}_{n\text{ summands}} = n.$$ The second sum is $$0+1+2+\cdots+(n-1)$$ and the third sum is $$0^2 + 1^2 + 2^2+\cdots + (n-1)^2.$$ There are well-known formulas to compute these sums: we have $$0+1+2+\cdots+(n-1) = \frac{(n-1)n}{2}$$ and $$0^2 + 1^2+\cdots+(n-1)^2 = \frac{(n-1)n(2n-1)}{6}.$$ So in the end we have $$\begin{align*} \text{LEFT}(n) &= \frac{2}{n^3}\left( 2n^2\sum_{i=0}^{n-1}1 - 6n\sum_{i=0}^{n-1}i + 4\sum_{i=0}^{n-1}i^2\right)\\ &= \frac{2}{n^3}\left( 2n^2(n) - 6n\frac{(n-1)n}{2} + 4\frac{(n-1)n(2n-1)}{6}\right)\\ &= \frac{4n^3}{n^3} - \frac{6n^2(n-1)}{n^3} + \frac{8(n-1)n(2n-1)}{6n^3}\\ &= 4 - \frac{6n-6}{n} + \frac{4(n-1)(2n-1)}{3n^2}\\ &= 4 - 6 + \frac{6}{n} + \frac{4(2n^2-3n + 1)}{3n^2}\\ &= -2 + \frac{6}{n} + \frac{8n^2}{3n^2} - \frac{12n}{3n^2} + \frac{4}{3n^2}\\ &= -2 + \frac{6}{n} + \frac{8}{3} - \frac{4}{n} + \frac{4}{3n^2}. \end{align*}$$

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    I do not know if I am doing a left or a right, the book does not specify.2012-04-14
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    I know Stewart's book; I'll wager dollars to donuts that the problem is clear, you just aren't understanding the instructions. You seem to believe that Calculus is drill, just like precalculus. It's not; it requires more skills than simple drill and memorization. And given your comments on how hard you found precalculus, I would guess that drill and memorization aren't your strong suit either. No wonder you have trouble with the subject (which is not easy to begin with)...2012-04-14
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    The instructions are to use the form of the defintion of the intregral given in theorem 4 to evaluate the integral. [4] Theorem says nothing about a side.2012-04-14
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    @Jordan: And so you show that you did not provide us with the relevant information in order to know what it is the book is asking you to do. As I say in my second sentence, I have to guess at what the problem might be, because you don't provide enough information to make it clear. I don't have Stewart's book in front of me, so I don't know what "Theorem 4" might or might not say, nor what you are expected to know at that stage of the book.2012-04-14
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    Well I have no idea how to type out a formula like this and I dont even really understand it so I was guessing it was a fairly common way to find an integral.2012-04-14
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    @Jordan: Given your difficulties digesting the material in this course, do you think you are in a good position to guess accurately what is important/common and what is not? If you can't figure out how to give the information, then you should have provided enough bibliographical information for someone to look it up and perhaps help you out by typing it. As in, "this is problem `xxx` in Section `Y.Z` of the `N`th edition of Stewart's `insert precise title since there are several versions of Stewart's calculus`." Can't help you until Monday, though, since my copies are at the office.2012-04-14
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    Just because I am horrible at math doesn't mean I am an idiot, I am guessing this is some fundemental definition of an integral otherwise it would have spanned two sections of my book.2012-04-14
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    I'm not saying you are an idiot: I'm saying that you are not in a position to judge accurately what information is relevant and what information is not *because you are not doing well in this course*. **I** am in no position to tell you what information may or may not be relevant on the dashboard of an airplane, so if I wanted help interpreting the dashboard, I would not try to guess. Because you routinely fail to provide enough information, we have to guess, and so you often get answers that are not to the point and only confuse you more. Like this answer.2012-04-14