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The limit

$\quad\quad \displaystyle \lim_{n\to\infty}\sum_{i=1}^{n} \frac{21 \cdot \frac{3 i}{n} + 18}{n} $

is the limit of a Riemann sum for a certain definite integral

$\quad\quad \displaystyle \int_a^b f(x)\, dx $

a = ?

b = ?

f(x) = ?

Attempt at solution:

So I know:

$ \displaystyle \sum_{i=1}^{n} f(a + i dx) dx $ = $ \int_a^b f(x)\, dx $

and that $x_i = a + i \cdot \frac {b-a}{n}$

So I will try to rewrite my riemann sum as convenient:

$\frac {3}{n} (7 \cdot \frac {3i}{n} + 6) $

so my $dx$ is $\frac {3}{n}$

since +6 is supposed to be my a, b is therefore 9 because $dx = \frac {b-a}{n}$

but that 7 inside the parenthesis spoils everything... so I can't say what's inside the parenthesis is equal to $x_i$

So what's next?

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3 Answers 3

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Divide through by 63. We then find that $f(a)=\frac{18}{63}$ by plugging in $i=0$. But then $f(x)=\frac{18}{63}+x$ because $f(a+idx)=f(a)+idx$. Hence $a=0$ and $b=1$. Therefore our integral is $63\int_{0}^{1} x+\frac{18}{63}dx$

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  • $\begingroup$ "by plugging in n=0." But you can't plug in 0 for n, since you will divide by 0 then. And how did you decide that a is 0 and b = 1? Why can't a = 1, b =2? $\endgroup$ Commented May 5, 2015 at 13:18
  • $\begingroup$ My apologies, you're very right, I meant $i=0$. I chose $a=0$ because it was convenient but my choice was totally arbitrary. The important part was that $f(a)=\frac{18}{63}$ and that $f(a+dx)=f(a)+dx$ which implies that $f(x)=x+c$, owing to the linearity. The function $f(x)=x-\frac{45}{63}$ with $a=1$ and $b=2$ is also perfectly valid. A quick check does indeed yield that $63\int_{0}^{1}x+\frac{18}{63}dx=63\int_{1}^{2} x-\frac{45}{63}dx$. If I'm to be entirely honest there is some handwaving done here; let me know if there's anything else that's unclear =). $\endgroup$ Commented May 5, 2015 at 16:43
  • $\begingroup$ I don't follow at all. I don't see how you got 18/63 $\endgroup$ Commented May 6, 2015 at 1:15
  • $\begingroup$ So we know that the sum $\sum_{i=1}^{\frac{b-a}{dx}} f(a+idx)dx=\int_{a}^{b} f(x)dx$, right? Well the given sum is $\lim_{n \to \infty} \sum_{i=1}^{n} \frac{\frac{63i}{n}+18}{n}=63\sum_{i=1}^{\frac{b-a}{dx}} (\frac{18}{63}+idx)dx$ but if $i=0$ then this implies that $f(a)=\frac{18}{63}$ since we've gotten rid of the $dx$ component in $f(a+idx)$. But then $f(a+idx)=\frac{18}{63}+idx$. That is that $f(x)=x+c$ while $f(a)=\frac{18}{63}$ and $b=a+1$. Everything else is arbitrary at that point as all integrals which satisfy these three conditions are equivalent. $\endgroup$ Commented May 6, 2015 at 1:33
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You were so close with $\frac {3}{n} (7 \cdot \frac {3i}{n} + 6)$. You can simply factor the $7$ outside the sum as follows:

Given the Riemann sum definition of the definite integral, $$\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(a + i \, \mathrm{d}x) \, \mathrm{d}x = \int_a^b f(x)\, \mathrm{d}x$$ we manipulate to find $f(x)$:

\begin{align} \lim_{n\to\infty}\sum_{i=1}^{n} \frac{21 \cdot \frac{3 i}{n} + 18}{n} &= \lim_{n\to\infty}\sum_{i=1}^{n} \left(6 + 7 i \cdot\frac{3}{n} \right)\frac{3}{n} \\ &= \lim_{n \rightarrow \infty} \sum_{i=1}^{n} 7\left(\frac{6}{7} + i \cdot\frac{3}{n} \right)\frac{3}{n} \\ &= 7\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \left(\frac{6}{7} + i \cdot\frac{3}{n} \right)\frac{3}{n}. \end{align}

So, we have $\mathrm{d}x=\frac{3}{n}$, $a=\frac{6}{7}$, $b=3+a=\frac{27}{7}$ and $f(x)=x$.

Hence, \begin{align} \lim_{n\to\infty}\sum_{i=1}^{n} \frac{21 \cdot \frac{3 i}{n} + 18}{n} &=7\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \left(\frac{6}{7} + i \cdot\frac{3}{n} \right)\frac{3}{n} \\ &= 7 \int_{\frac{6}{7}}^{\frac{27}{7}} x \,\mathrm{d}x \\ &= 7 \left[\frac{1}{2} x^2 \right]_{\frac{6}{7}}^{\frac{27}{7}} \\ &= 7 \left[\frac{729}{98} - \frac{36}{98} \right] \\ &= 49.5 \end{align}

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You don't need to adjust your boundaries for the integral. The easiest way is to choose arbitrarily $a=0$ and $b=1$ so that $\Delta x = 1/n$ and $x^{\ast}$ is $i/n$. See the picture: [1]

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