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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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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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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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