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I am suppose to use Theorem 4 in my books (just the limit as n approaches infinity of a summation representing the function).

I am trying to evaluate

$ \int_2^5 (4-2x) dx$

I really have no idea what to do. I know $dx$ is $3/n$ and that the summation should look something like $\frac{n(n+1)}{2}$ after using the rules of wizardry that I memorized from earlier in the chapter.

As far as what to do now I tried to put that back into the equation and got nothing logical.

$4-2 n(n+1)$ doesn't really make any sense to me.

From here I do not know what to do.

  • 1
    Downvoter should explain himself. This is bad MSE behavior in my opinion. A downvote should be instructive ; explain what you dislike about this question. Personally I think everything's okay, so this is a bad downvote.2012-04-13

1 Answers 1

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Let’s build up the Riemann sum first. The interval is $[2,5]$, so its length is $3$, and when you divide it into $n$ equal subintervals, each will be of length $\frac3n$, so $\Delta x$ (not $dx$) is indeed $\frac3n$. The ends of the subintervals $-$ the $x_k$’s $-$ are $2+\frac3n,2+2\left(\frac3n\right)$, and so on, with $x_k=2+\frac{3k}n$. Thus, your $n$-th Riemann sum, $R_n$, is

$\begin{align*} R_n=\sum_{k=1}^n(4-2x_k)\frac3n&=\sum_{k=1}^n\left(4-2\left(2+\frac{3k}n\right)\right)\frac3n\\ &=\sum_{k=1}^n\left(4-4-\frac{6k}n\right)\frac3n\\ &=\sum_{k=1}^n\left(-\frac{6k}n\right)\frac3n\\ &=\sum_{k=1}^n\left(-\frac{18k}{n^2}\right)\;. \end{align*}$

Now that’s just $-\frac{18}{n^2}-\frac{18\cdot 2}{n^2}-\frac{18\cdot 3}{n^2}-\ldots-\frac{18n}{n^2}\;,$

with a factor of $-\dfrac{18}{n^2}$ in every term that we can factor out to get $-\frac{18}{n^2}(1+2+3+\ldots+n)=-\frac{18}{n^2}\sum_{k=1}^nk\;.$ As you said in the question, you know that $\sum_{k=1}^nk$, the sum of the first $n$ positive integers, is $\frac{n(n+1)}2$, so $\begin{align*}R_n&=-\frac{18}{n^2}\sum_{k=1}^nk=-\frac{18}{n^2}\cdot\frac{n(n+1)}2\\&=-\frac{9(n+1)}n=-9\cdot\frac{n+1}n\\&=-9\left(1+\frac1n\right)\;.\end{align*}$

Finally, $\begin{align*}\int_2^5(4-2x)dx&=\lim_{n\to\infty}R_n\\&=\lim_{n\to\infty}-9\left(1+\frac1n\right)\\&=-9\lim_{n\to\infty}\left(1+\frac1n\right)\;.\end{align*}$

Now, what’s $\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)$?

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    Yes, it does: $\sum_{k=1}^nk=\frac{n(n+1)}2$, and $\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}6$.2012-04-13