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 \(\displaystyle{R}_{{{n}}}\) of the function \(\displaystyle{f{{\left({x}\right)}}}={4}-{x}^{{{2}}}\) on the interval \(\displaystyle{\left[{0},\ {2}\right]}\) using n subintervals of equal length, and calculate the definite integral \(\displaystyle{\int_{{{0}}}^{{{2}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\) as the limit of \(\displaystyle{R}_{{{n}}}\) at \(\displaystyle{n}\rightarrow\infty\).
(Reminder: \(\displaystyle{\sum_{{{k}={1}}}^{{{n}}}}{k}={n}\frac{{{n}+{1}}}{{2}},\ {\sum_{{{k}={1}}}^{{{n}}}}{k}^{{{2}}}={n}{\left({n}+{1}\right)}\frac{{{2}{n}+{1}}}{{6}}{)}\)
b) Use the Fundamental Theorem of Calculus to calculate the derivative of \(\displaystyle{F}{\left({x}\right)}={\int_{{{e}^{{-{x}}}}}^{{{x}}}}{\ln{{\left({t}^{{{2}}}+{1}\right)}}}{\left.{d}{t}\right.}\)

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Expert Answer

lamusesamuset
Answered 2021-08-21 Author has 18204 answers
Step 1
a) Given: \(\displaystyle{f{{\left({x}\right)}}}={4}-{x}^{{{2}}},\ {a}={0},\ {b}={2},\ \Delta{x}=\frac{{{b}-{a}}}{{n}}={\left({2}-{0}\right)}_{{{n}}}={\left(\frac{{2}}{{n}}\right)}\)
\(\displaystyle{x}_{{{i}}}={a}+{i}\Delta{x}={0}+{i}{\left(\frac{{2}}{{n}}\right)}={\left({2}\frac{{i}}{{n}}\right)}\)
\(\displaystyle{f{{\left({x}_{{{i}}}\right)}}}={4}-{\left({2}\frac{{i}}{{n}}\right)}^{{{2}}}\)
\(\displaystyle{R}_{{{n}}}={\sum_{{{i}={1}}}^{{{n}}}}{f{{\left({x}_{{{i}}}\right)}}}\Delta{x}\)
\(\displaystyle{R}_{{{n}}}={\sum_{{{i}={1}}}^{{{n}}}}{\left({4}-{\left({2}\frac{{i}}{{n}}\right)}^{{{2}}}\right)}{\left(\frac{{2}}{{n}}\right)}\)
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}=\lim_{{{n}\geq\infty}}{R}_{{{n}}}\)
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}=\lim_{{{n}\geq\infty}}{\sum_{{{i}={1}}}^{{{n}}}}{\left({4}-{\left({2}\frac{{i}}{{n}}\right)}^{{{2}}}\right)}{\left(\frac{{2}}{{n}}\right)}\)
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}=\lim_{{{n}\geq\infty}}{\sum_{{{i}={1}}}^{{{n}}}}{\left({1}-{\left(\frac{{i}^{{{2}}}}{{n}^{{{2}}}}\right)}\right)}{\left(\frac{{8}}{{n}}\right)}\)
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}=\lim_{{{n}\geq\infty}}{\sum_{{{i}={1}}}^{{{n}}}}{8}{\left({\left(\frac{{1}}{{n}}\right)}-{\left(\frac{{i}^{{{2}}}}{{n}^{{{3}}}}\right)}\right)}\)
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}=\lim_{{{n}\geq\infty}}{8}{\left({\left(\frac{{n}}{{n}}\right)}-{\left({n}{\left({n}+{1}\right)}\frac{{{2}{n}+{1}}}{{6}}{n}^{{{3}}}\right)}\right)}\)
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}=\lim_{{{n}\geq\infty}}{8}{\left({1}-{\left({1}{\left({1}+{\left(\frac{{1}}{{n}}\right)}\right)}\frac{{{2}+{\left(\frac{{1}}{{n}}\right)}}}{{6}}\right)}\right)}\)
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={8}\times{\left({1}-{\left({1}{\left({1}+{0}\right)}\frac{{{2}+{0}}}{{6}}\right)}\right)}\)
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={8}\times{\left({1}-{\left(\frac{{2}}{{6}}\right)}\right)}\)
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={8}\times{\left(\frac{{2}}{{3}}\right)}\)
\(\displaystyle{\int_{{{0}}}^{{{2}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={\left(\frac{{16}}{{3}}\right)}\)
Step 2
b) \(\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\int_{{{p}{\left({x}\right)}}}^{{{q}{\left({x}\right)}}}}{f{{\left({t}\right)}}}{\left.{d}{t}\right.}={f{{\left({q}{\left({x}\right)}\right)}}}\times{q}'{\left({x}\right)}-{f{{\left({p}{\left({x}\right)}\right)}}}\times{p}'{\left({x}\right)}\)
\(\displaystyle{F}{\left({x}\right)}={\int_{{{e}^{{-{x}}}}}^{{{x}}}}{\ln{{\left({t}^{{{2}}}+{1}\right)}}}{\left.{d}{t}\right.}\)
\(\displaystyle{F}'{\left({x}\right)}={\left({\ln{{\left({x}^{{{2}}}+{1}\right)}}}\right)}\times{\left({1}\right)}-{\left({\ln{{\left({\left({e}^{{-{x}}}\right)}^{{{2}}}+{1}\right)}}}\right)}\times{\left(-{e}^{{-{x}}}\right)}\)
\(\displaystyle{F}'{\left({x}\right)}={\ln{{\left({x}^{{{2}}}+{1}\right)}}}+{e}^{{-{x}}}{\ln{{\left({e}^{{-{2}{x}}}+{1}\right)}}}\)
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