# Integral Calculus 2

I have to find a formula...like n(n+1)(2n+1)/6 , but I cannot find it.

$f(x) = \frac{-x^2}4 + 6$ over the points $[0,4]$

The summation inside the brackets is $R_n$ which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

Calculate $R_n$ for the function with the answer being the function of $n$ without any summation signs.

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@user5955: Welcome to MSE! What have you tried? This information helps the MSE Community in understanding your issues and where are you confused? How many rectangles are you supposed to use? Is this homework? If so, please tag it as such. Regards –  Amzoti Jan 22 '13 at 19:21

If I have understood correctly, you want to evaluate: $$A=\lim_{n\to+\infty}\sum_1^nf(x_i)\Delta x, ~~f(x)=\frac{-x^2}{4}+6$$ Set $\Delta x=\frac{4-0}{n}$ in which we divide the closed interval $[0,4]$ into $n$ sub intervals, each of length $\Delta x$. In fact $$x_0=0, x_1=0+\Delta x, x_2=0+2\Delta x,...,x_{n-1}=0+(n-1)\Delta x, x_n=4$$ Since, the function $f(x)$ is decreasing on the interval, so the absolute min value of $f$ on the $i$th subinterval $[x_{i-1},x_i]$ is $f(x_i)$. But $f(x_i)=f(0+i\Delta x)$ as you see above and then $$f(x_i)=-\frac{(i\Delta x)^2}{4}+6=-\frac{i^2\Delta^2x}{4}+6$$

Now, let's try to find the above summation: $$A=\lim_{n\to+\infty}\sum_1^nf(x_i)\Delta x=\lim_{n\to+\infty}\sum_1^n\left(-\frac{i^2\Delta^2x}{4}+6\right)\Delta x$$ wherein $\Delta x=\frac{4}{n}$.

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So what is Rn = ?? –  user59255 Jan 22 '13 at 19:35
I know, I need to know the function. What is it and how do you find it? –  user59255 Jan 22 '13 at 19:40
@user59255: Isn't the function you noted, our function? Do you know anything about the Riemannian sum for an definite integral? For finding that summation, we need a continuous function considered on an interval and you have one. $f(x)=6-x^2/4$ –  Babak S. Jan 22 '13 at 19:45
So, Rn = 17.84 and lim n--> inf of Rn = 56/3 ?? –  user59255 Jan 23 '13 at 3:58
@user59255: yes. see the fig above. –  Babak S. Jan 23 '13 at 5:33

The right endpoints of the subintervals are $4k/n$ for $1 \le k \le n$, and the common width of subintervals is $4/n$. So the value of $R_n$, the right hand sum, is $$\sum_{k=1}^n [ (-1/4) \cdot (4k/n)^2 +6]\cdot(4/n)$$ The constant +6 added $n$ times gives $6n$, which is then multiplied by $(4/n)$ so contributes $+24$ to $R_n$. The remaining part of the sum is $$(-1/4)\cdot (4^2/n^2) \cdot (4/n) \cdot \sum_{k=1}^n k^2.$$ From here just use your formula for the sum of the squares of the first $n$ positive integers, plug in and simplify, not forgetting to add the $+24$.

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