 # a) Write the sigma notation formula for the right Riemann sum R_{n} of the function f(x)=4-x^{2} on the interval [0,\ 2] using n subintervals of equal length, and calculate the definite integral \int_{0}^{2}f(x) dx as the limit of R_{n} at n\rightarrow\infty.(Reminder: \sum_{k=1}^{n}k=n(n+1)/2,\ \sum_{k=1}^{n}k^{2}=n(n+1)(2n+1)/6) djeljenike 2021-08-20 Answered
a) Write the sigma notation formula for the right Riemann sum ${R}_{n}$ of the function $f\left(x\right)=4-{x}^{2}$ on the interval using n subintervals of equal length, and calculate the definite integral ${\int }_{0}^{2}f\left(x\right)dx$ as the limit of ${R}_{n}$ at $n\to \mathrm{\infty }$.
(Reminder:
b) Use the Fundamental Theorem of Calculus to calculate the derivative of $F\left(x\right)={\int }_{{e}^{-x}}^{x}\mathrm{ln}\left({t}^{2}+1\right)dt$
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Step 1
a) Given:
${x}_{i}=a+i\mathrm{\Delta }x=0+i\left(\frac{2}{n}\right)=\left(2\frac{i}{n}\right)$
$f\left({x}_{i}\right)=4-{\left(2\frac{i}{n}\right)}^{2}$
${R}_{n}=\sum _{i=1}^{n}f\left({x}_{i}\right)\mathrm{\Delta }x$
${R}_{n}=\sum _{i=1}^{n}\left(4-{\left(2\frac{i}{n}\right)}^{2}\right)\left(\frac{2}{n}\right)$
${\int }_{0}^{2}f\left(x\right)dx=\underset{n\ge \mathrm{\infty }}{lim}{R}_{n}$
${\int }_{0}^{2}f\left(x\right)dx=\underset{n\ge \mathrm{\infty }}{lim}\sum _{i=1}^{n}\left(4-{\left(2\frac{i}{n}\right)}^{2}\right)\left(\frac{2}{n}\right)$
${\int }_{0}^{2}f\left(x\right)dx=\underset{n\ge \mathrm{\infty }}{lim}\sum _{i=1}^{n}\left(1-\left(\frac{{i}^{2}}{{n}^{2}}\right)\right)\left(\frac{8}{n}\right)$
${\int }_{0}^{2}f\left(x\right)dx=\underset{n\ge \mathrm{\infty }}{lim}\sum _{i=1}^{n}8\left(\left(\frac{1}{n}\right)-\left(\frac{{i}^{2}}{{n}^{3}}\right)\right)$
${\int }_{0}^{2}f\left(x\right)dx=\underset{n\ge \mathrm{\infty }}{lim}8\left(\left(\frac{n}{n}\right)-\left(n\left(n+1\right)\frac{2n+1}{6}{n}^{3}\right)\right)$