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Riemann Integral. Simpson's Rule. Approximation - nerjik.com

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Riemann Integral. Simpson's Rule. Approximation

Riemann Integral. Simpson's Rule. Approximation

Almaz Khusnutdinov

Almaz Khusnutdinov•Math enthusiast, love coding and mathematics

Published 9/22/2025

Evaluate Series Applying Riemann Integral

Problem 1.1

\text{Evaluate:}\\ \text{(a) } \displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\sin\!\frac{k\pi}{n},\qquad \text{(b) } \displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\tan\!\frac{k\pi}{3n}.\\[8pt] \textbf{Answer: } (a)\ \dfrac{2}{\pi};\quad (b)\ \dfrac{3}{\pi}\ln 2.\\[8pt] \textbf{Solution: }\\ \text{(a) Let }x_k=\tfrac{k\pi}{n},\ \Delta x=\tfrac{\pi}{n}.\ \frac{1}{n}\sum_{k=1}^{n}\sin\!\frac{k\pi}{n} =\frac{1}{\pi}\sum_{k=1}^{n}\sin(x_k)\,\Delta x\ \xrightarrow[n\to\infty]{}\ \frac{1}{\pi}\int_{0}^{\pi}\sin x\,dx=\frac{2}{\pi}.\\[6pt] \text{(b) Let }y_k=\tfrac{k\pi}{3n},\ \Delta y=\tfrac{\pi}{3n}.\ \frac{1}{n}\sum_{k=1}^{n}\tan\!\frac{k\pi}{3n} =\frac{3}{\pi}\sum_{k=1}^{n}\tan(y_k)\,\Delta y\ \xrightarrow[n\to\infty]{}\ \frac{3}{\pi}\int_{0}^{\pi/3}\tan x\,dx\\ =\frac{3}{\pi}\big[-\ln(\cos x)\big]_{0}^{\pi/3}=\frac{3}{\pi}\ln 2.

Problem 1.2

\text{Evaluate:}\\ \text{(a) } \displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\cos^{2}\!\frac{k\pi}{n},\qquad \text{(b) } \displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\sec^{2}\!\frac{k\pi}{4n}.\\[6pt] \textbf{Answer: } (a)\ \dfrac{1}{2};\quad (b)\ \dfrac{4}{\pi}.

Problem 1.3

\text{Evaluate:}\\ \text{(a) } \displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\!\left(\sin\!\frac{k\pi}{2n}\right)\!\left(\cos\!\frac{k\pi}{2n}\right),\quad \text{(b) } \displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\sin^{2}\!\frac{k\pi}{2n}.\\[6pt] \textbf{Answer: } (a)\ \dfrac{1}{\pi};\quad (b)\ \dfrac{1}{2}.

Problem 1.4

\text{Evaluate:}\\ \text{(a) } \displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\big|\sin\!\frac{2\pi k}{n}\big|,\qquad \text{(b) } \displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\cos\!\frac{2\pi k}{n}.\\[6pt] \textbf{Answer: } (a)\ \dfrac{2}{\pi};\quad (b)\ 0.

Problem 1.5

\text{Evaluate:}\\ \text{(a) } \displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{\sin\!\frac{k\pi}{n}}{2+\cos\!\frac{k\pi}{n}},\qquad \text{(b) } \displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1-\cos\!\frac{k\pi}{2n}}{1+\cos\!\frac{k\pi}{2n}}.\\[6pt] \textbf{Answer: } (a)\ \dfrac{1}{\pi}\ln 3;\quad (b)\ \dfrac{4}{\pi}-1.

Integral Mean Value Theorem

Problem 2.1

\text{Let }f\text{ be continuous on }[a,b].\ \text{Prove that there exists }c\in[a,b]\text{ with } \int_{a}^{b}f(x)\,dx=(b-a)f(c).\\[12pt] \textbf{Detailed solution:}\\[6pt] \text{Let }m=\min_{x\in[a,b]}f(x),\quad M=\max_{x\in[a,b]}f(x).\text{ Continuity on the compact interval }[a,b]\\ \text{ensures }m,M\text{ exist and }m\le f(x)\le M\ \forall x\in[a,b].\text{ Integrating these inequalities gives }\\[6pt] m(b-a)=\int_{a}^{b}m\,dx\le\int_{a}^{b}f(x)\,dx\le\int_{a}^{b}M\,dx=M(b-a).\\[6pt] \text{Divide by }(b-a)>0\text{ to obtain }m\le \frac{1}{b-a}\int_{a}^{b}f(x)\,dx\le M.\\[6pt] \text{But }f([a,b])=[m,M]\text{ because }f\text{ is continuous (the image of a connected set is connected). Hence} \frac{1}{b-a}\int_{a}^{b}f(x)\,dx\in[m,M]=f([a,b]).\\[6pt] \text{Therefore there exists }c\in[a,b]\text{ with }f(c)=\dfrac{1}{b-a}\int_{a}^{b}f(x)\,dx, \text{ i.e. }\int_{a}^{b}f(x)\,dx=(b-a)f(c).\!

Problem 2.2

\text{Let }f,g\text{ be continuous on }[a,b]\text{ with }g(x)\ge 0\text{ on }[a,b]\text{ and }g\not\equiv 0. \ \text{Show there exists }c\in[a,b]\text{ with }\\[6pt] \int_{a}^{b}f(x)g(x)\,dx=f(c)\int_{a}^{b}g(x)\,dx.\\[12pt] \textbf{Proof outline:}\\[6pt] \text{Let }m=\min_{[a,b]}f,\ M=\max_{[a,b]}f.\ \text{Then }m g(x)\le f(x)g(x)\le M g(x)\ \forall x.\\[6pt] \text{Integrate to get }m\int_{a}^{b}g(x)\,dx\le\int_{a}^{b}f(x)g(x)\,dx\le M\int_{a}^{b}g(x)\,dx.\\[6pt] \text{If }I:=\int_{a}^{b}g(x)\,dx>0\text{ then divide by }I\text{ to see }m\le\frac{1}{I}\int_{a}^{b}f g\le M.\\[6pt] \text{By continuity of }f\text{ its range on }[a,b]\text{ is }[m,M],\text{ so some }c\in[a,b]\text{ satisfies } f(c)=\frac{1}{I}\int_{a}^{b}f g.\\[6pt] \text{Hence }\int_{a}^{b}f g=f(c)\int_{a}^{b}g.\!

Problem 2.3

\text{(a) Let }f(x)=x^{2}\text{ on }[1,3].\quad \text{(b) Compute }A=\dfrac{1}{3-1}\int_{1}^{3}x^{2}\,dx.\quad \text{(c) Find }c\in[1,3]\text{ with }f(c)=A.\\[12pt] \textbf{Answer: }A=\dfrac{13}{3},\qquad c=\sqrt{\dfrac{13}{3}}\ (\approx2.081665\ldots).

Problem 2.4

\text{(a) Let }f(x)=\sin x\text{ on }\big[\tfrac{\pi}{6},\tfrac{5\pi}{6}\big].\quad \text{(b) Compute }A=\dfrac{1}{\tfrac{5\pi}{6}-\tfrac{\pi}{6}}\int_{\pi/6}^{5\pi/6}\sin x\,dx.\quad \text{(c) Find }c\in\big[\tfrac{\pi}{6},\tfrac{5\pi}{6}\big]\text{ with }\sin c=A.\\[12pt] \textbf{Answer: }A=\dfrac{3\sqrt{3}}{2\pi},\qquad c=\arcsin\!\bigg(\dfrac{3\sqrt{3}}{2\pi}\bigg)\ (\in(\tfrac{\pi}{6},\tfrac{\pi}{2})).

Problem 2.5

\text{(a) Let }f(x)=e^{-x}\text{ on }[0,2].\quad \text{(b) Compute }A=\dfrac{1}{2-0}\int_{0}^{2}e^{-x}\,dx.\quad \text{(c) Find }c\in[0,2]\text{ with }e^{-c}=A.\\[12pt] \textbf{Answer: }A=\tfrac{1-e^{-2}}{2},\qquad c=-\ln\!\bigg(\tfrac{1-e^{-2}}{2}\bigg)\ (\approx0.838\ldots).

Problem 2.6

\text{(a) Let }f(x)=\dfrac{1}{1+x^{2}}\text{ on }[0,1].\quad \text{(b) Compute }A=\dfrac{1}{1-0}\int_{0}^{1}\dfrac{1}{1+x^{2}}\,dx.\quad \text{(c) Find }c\in[0,1]\text{ with }\dfrac{1}{1+c^{2}}=A.\\[12pt] \textbf{Answer: }A=\dfrac{\pi}{4},\qquad c=\sqrt{\dfrac{4}{\pi}-1}\ (\in(0,1)).

Differentiation of Integral-Based Expression

Problem 3.1

\text{Let }f\text{ be continuous on }[a,b]\text{ and define }F(x)=\displaystyle\int_{a}^{x}f(t)\,dt\text{ for }x\in[a,b].\\[4pt] \text{Show that }F\text{ is differentiable on }(a,b)\text{ and }F'(x)=f(x).\\[10pt] \textbf{Answer (outline): }\\ \text{Fix }x\in(a,b)\text{ and consider }\dfrac{F(x+h)-F(x)}{h}=\dfrac{1}{h}\int_{x}^{x+h}f(t)\,dt.\\ \text{Use continuity of }f\text{ at }x\text{ to bound }f(t)\text{ between }f(x)\pm\varepsilon\text{ on a small interval around }x,\\ \text{apply the squeeze theorem to the difference quotient, and conclude }F'(x)=f(x).

Problem 3.2

\text{Let }f\text{ be continuous on an interval containing }[a,b]\text{ and let }g\text{ be differentiable.}\\ \text{Define }H(x)=\displaystyle\int_{a}^{g(x)}f(t)\,dt.\ \text{Show that }H'(x)=f(g(x))g'(x).\\[10pt] \textbf{Answer (outline): }\\ \text{Write }H(x+h)-H(x)=\int_{g(x)}^{g(x+h)}f(t)\,dt.\ \text{Express this as }(g(x+h)-g(x))\cdot \text{(average of }f\text{ on that interval).}\\ \text{Use continuity of }f\text{ at }g(x)\text{ to control the average value, and the differentiability of }g\text{ to control }g(x+h)-g(x).\\ \text{Take the limit as }h\to0\text{ to obtain }H'(x)=f(g(x))g'(x).

Problem 3.3

\text{(a) Let }F(x)=\displaystyle\int_{0}^{x^{2}}\bigl(1+t^{3}\bigr)\,dt.\ \text{Find }F'(x).\\[4pt] \text{(b) Let }G(x)=\displaystyle\int_{x}^{1}\sqrt{1+t^{2}}\,dt.\ \text{Find }G'(x).\\[10pt] \textbf{Answer: }(a)\ F'(x)=2x\bigl(1+x^{6}\bigr).\quad (b)\ G'(x)=-\sqrt{1+x^{2}}.

Problem 3.4

\text{(a) Let }H(x)=\displaystyle\int_{1}^{\sin x}\frac{\ln t}{t}\,dt.\ \text{Find }H'(x).\\[4pt] \text{(b) Let }K(x)=\displaystyle\int_{x^{2}}^{x^{3}}e^{-t^{2}}\,dt.\ \text{Find }K'(x).\\[10pt] \textbf{Answer: }(a)\ H'(x)=\dfrac{\ln(\sin x)}{\sin x}\cos x.\quad (b)\ K'(x)=3x^{2}e^{-x^{6}}-2x e^{-x^{4}}.

Problem 3.5

\text{(a) Let }y_{1}(x)=\displaystyle\int_{0}^{x}\cos(t^{2})\,dt.\ \text{Find }y_{1}'(x).\\[4pt] \text{(b) Let }y_{2}(x)=\displaystyle\int_{0}^{x}\frac{1}{1+t^{4}}\,dt.\ \text{Find }y_{2}'(x).\\[10pt] \textbf{Answer: }(a)\ y_{1}'(x)=\cos(x^{2}).\quad (b)\ y_{2}'(x)=\dfrac{1}{1+x^{4}}.

Problem 3.6

\text{(a) Let }f_{1}(x)=\displaystyle\int_{-2}^{x}\frac{t^{2}+1}{t^{2}+4}\,dt.\ \text{Find }f_{1}'(x).\\[4pt] \text{(b) Let }f_{2}(x)=\displaystyle\int_{-x}^{x}e^{t^{2}}\,dt.\ \text{Find }f_{2}'(x).\\[10pt] \textbf{Answer: }(a)\ f_{1}'(x)=\dfrac{x^{2}+1}{x^{2}+4}.\quad (b)\ f_{2}'(x)=2e^{x^{2}}.

Problem 3.7

\text{(a) Let }F(x)=x^{2}\displaystyle\int_{0}^{x}\sqrt{1+t^{4}}\,dt.\ \text{Find }F'(x).\\[4pt] \text{(b) Let }G(x)=\displaystyle\int_{x}^{x^{2}}\ln\bigl(1+t^{2}\bigr)\,dt.\ \text{Find }G'(x).\\[10pt] \textbf{Answer: }(a)\ F'(x)=2x\int_{0}^{x}\sqrt{1+t^{4}}\,dt+x^{2}\sqrt{1+x^{4}}.\\[4pt] (b)\ G'(x)=2x\ln\bigl(1+x^{4}\bigr)-\ln\bigl(1+x^{2}\bigr).

Trapezoidal Rule

Problem 4.1

\text{Composite trapezoidal rule on }[0,1]\text{ for }f(x)=x^2.\ \ h=\tfrac{1}{n}.\\[4pt] \text{(a) Write }T_n=h\Big[\tfrac{f(0)+f(1)}{2}+\sum_{k=1}^{n-1}f(kh)\Big].\\ \text{(b) Find a closed form for }T_n\text{ and the exact error }E_n=T_n-\int_{0}^{1}x^2\,dx.\\ \text{(c) Smallest }n\text{ with }|E_n|<10^{-3}.\\[8pt] \textbf{Answer: }T_n=\dfrac{2n^2+1}{6n^2},\quad \int_{0}^{1}x^2\,dx=\dfrac{1}{3},\quad E_n=\dfrac{1}{6n^2},\quad n_{\min}=13.\\[8pt] \textbf{Solution: }T_n=\tfrac{1}{n}\!\left[\tfrac{0+1}{2}+\sum_{k=1}^{n-1}\!\Big(\tfrac{k}{n}\Big)^2\right] =\tfrac{1}{n}\!\left[\tfrac{1}{2}+\tfrac{1}{n^2}\sum_{k=1}^{n-1}\!k^2\right].\\ \sum_{k=1}^{n-1}\!k^2=\tfrac{(n-1)n(2n-1)}{6}\ \Rightarrow\ T_n=\tfrac{1}{n}\!\left[\tfrac{1}{2}+\tfrac{(n-1)(2n-1)}{6n}\right]=\tfrac{2n^2+1}{6n^2}.\\ \int_{0}^{1}x^2dx=\tfrac{1}{3}\ \Rightarrow\ E_n=T_n-\tfrac{1}{3}=\tfrac{1}{6n^2}.\ |E_n|<10^{-3}\ \Rightarrow\ n^2>\tfrac{1}{6\cdot10^{-3}}=166.\overline{6}\ \Rightarrow\ n_{\min}=13.

Problem 4.2

\text{Composite trapezoidal rule on }[0,1]\text{ for }f(x)=x^3.\ \ h=\tfrac{1}{n}.\\[4pt] \text{(a) }T_n=h\Big[\tfrac{f(0)+f(1)}{2}+\sum_{k=1}^{n-1}f(kh)\Big].\quad \text{(b) }E_n=T_n-\int_{0}^{1}x^3dx.\quad \text{(c) Smallest }n\text{ with }|E_n|<10^{-3}.\\[8pt] \textbf{Answer: }T_n=\dfrac{n^2+1}{4n^2},\ \ \int_0^1 x^3dx=\dfrac{1}{4},\ \ E_n=\dfrac{1}{4n^2},\ \ n_{\min}=16.\\[8pt]

Problem 4.3

\text{Composite trapezoidal rule on }[0,2]\text{ for }f(x)=x^2.\ \ h=\tfrac{2}{n}.\\[4pt] \text{(a) Compute }T_n.\quad \text{(b) Exact }E_n=T_n-\int_0^2 x^2dx.\quad \text{(c) Smallest }n\text{ with }|E_n|<10^{-3}.\\[8pt] \textbf{Answer: }T_n=\dfrac{8}{3}+\dfrac{4}{3n^2},\ \ \int_0^2 x^2dx=\dfrac{8}{3},\ \ E_n=\dfrac{4}{3n^2},\ \ n_{\min}=37.\\[8pt]

Problem 4.4

\text{Composite trapezoidal rule on }[0,\pi]\text{ for }f(x)=\cos x.\ \ h=\tfrac{\pi}{n}.\\[4pt] \text{(a) Compute }T_n.\quad \text{(b) Exact }E_n=T_n-\int_{0}^{\pi}\cos x\,dx.\quad \text{(c) Smallest }n\text{ with }|E_n|<10^{-3}.\\[10pt] \textbf{Answer: }T_n=0,\ \ \int_{0}^{\pi}\cos x\,dx=0,\ \ E_n=0,\ \ n_{\min}=1.\\[10pt]

Problem 4.5

\text{Composite trapezoidal rule on }[0,1]\text{ for }f(x)=e^{x}.\ \ h=\tfrac{1}{n}.\\[4pt] \text{(a) Compute }T_n.\quad \text{(b) Exact }E_n=T_n-(e-1).\quad \text{(c) Smallest }n\text{ with }|E_n|<10^{-3}.\\[8pt] \textbf{Answer: }T_n=\dfrac{1}{n}\!\left[\dfrac{1+e}{2}+\dfrac{e-e^{1/n}}{e^{1/n}-1}\right],\ E_n=T_n-(e-1),\ n_{\min}=16.\\[8pt]

Midpoint Rule

Problem 5.1

\text{Use the midpoint rule with }n=5\text{ subintervals on }[0,1]\text{ to approximate }\int_{0}^{1}x^{2}\,dx.\ \text{Answer: }0.33\text{ (approx).}

Problem 5.2

\text{Use the midpoint rule with }n=4\text{ subintervals on }[0,2]\text{ to approximate }\int_{0}^{2}e^{x}\,dx.\ \text{Answer: }6.3230\text{ (approx).}

Problem 5.3

\text{Use the midpoint rule with }n=6\text{ subintervals on }[0,\pi]\text{ to approximate }\int_{0}^{\pi}\sin x\,dx.\ \text{Answer: }2.0230\text{ (approx).}

Problem 5.4

\text{Use the midpoint rule with }n=4\text{ subintervals on }\bigl[0,\tfrac{\pi}{2}\bigr]\text{ to approximate }\int_{0}^{\pi/2}\cos x\,dx.\ \text{Answer: }1.0065\text{ (approx).}

Problem 5.5

\text{Use the midpoint rule with }n=4\text{ subintervals on }[0,2]\text{ to approximate }\int_{0}^{2}\frac{1}{1+x^{2}}\,dx.\ \text{Answer: }1.1088\text{ (approx).}

Simpson's Approximation

Problem 6.1

\text{Use Simpson's rule with }n=4\text{ equal subintervals on }[0,2]\text{ to approximate } \displaystyle\int_{0}^{2}x^{2}\,dx.\ \textbf{Answer: } \dfrac{8}{3}\approx 2.67.

Problem 6.2

\text{Use Simpson's rule with }n=4\text{ equal subintervals on }[0,1]\text{ to approximate } \displaystyle\int_{0}^{1}e^{x}\,dx.\ \textbf{Answer: } \approx 1.71832.

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