Exercises — Riemann sums — left, right, midpoint; formal definition of definite integral

Hum kya kar rahe hain: bas symbols decode kar rahe hain — abhi koi area computation nahi. $\Delta x=\frac{b-a}{n}=\frac{6-2}{8}=\frac{1}{2}.$ Grid points $x_i=a+i\Delta x=2+\frac{i}{2}$: $x_0=2,x_1=2.5,x_2=3,x_3=3.5,x_4=4.$ Teesra subinterval $[x_2,x_3]=[3,3.5]$ hai. Uska midpoint sample point hai $x_3^{\ast }=\frac{x_2+x_3}{2}=\frac{3+3.5}{2}=\mathrm{3.25.}$ Answer: $\Delta x=\frac{1}{2}$; grid shuru hoti hai $2,2.5,3,3.5,4$ se; teesra midpoint $x_3^{\ast }=3.25$.

Hum kya kar rahe hain: sampling location ko increasing $f$ ke liye over/under behaviour se connect kar rahe hain. Panels ko position se padho, colour se nahi.

• Leftmost panel — Left rule. Har rectangle apni strip ke left edge pe height leta hai. Ek rising curve pe left edge us strip ka sabse neecha point hota hai, isliye har rectangle curve ke neeche baithta hai → underestimate.
• Rightmost panel — Right rule. Har rectangle right edge leta hai, ek rising strip ka sabse uocha point → rectangles curve ke *upar nikalte hain → overestimate.
• Centre panel — Midpoint rule. Height strip ke centre pe li jaati hai; upar wala sliver jo bahar nikalta hai woh roughly neeche wale missing sliver ko cancel kar deta hai → sach ke sabse kareeb. Answer: Leftmost = Left = under; Rightmost = Right = over; Centre = Midpoint = best. (Figure mein colours — burnt orange, plum, teal — bas inhe decorate karte hain; position aur edge-choice hi inhe identify karti hai.)

Setup: $\Delta x=\frac{2-0}{4}=0.5$, grid $x_i=0.5i$: $0,0.5,1,1.5,2$.

Left $x_0,x_1,x_2,x_3=0,0.5,1,1.5$ pe sample karta hai (last point drop karo): $L_4=(f(0)+f(0.5)+f(1)+f(1.5))\Delta x=(0+0.25+1+2.25)(0.5)=3.5\times 0.5=\mathrm{1.75.}$ Right $x_1,x_2,x_3,x_4=0.5,1,1.5,2$ pe sample karta hai (first point drop karo): $R_4=(0.25+1+2.25+4)(0.5)=7.5\times 0.5=\mathrm{3.75.}$ Midpoint $0.25,0.75,1.25,1.75$ pe sample karta hai: $M_4=(0.0625+0.5625+1.5625+3.0625)(0.5)=5.25\times 0.5=\mathrm{2.625.}$ Ordering check karo: $f$ increasing hai, toh $L_4=1.75<M_4=2.625<R_4=3.75$, aur sach wala area ${\int }_0^2{x}^{2}dx=\frac{8}{3}\approx 2.667$ hai. Midpoint sabse kareeb hai. ✓

Setup: $\Delta x=\frac{4-0}{4}=1$, grid $0,1,2,3,4$. Right $x_1,\dots ,x_4=1,2,3,4$ pe sample karta hai. $R_4=(f(1)+f(2)+f(3)+f(4))(1)=(-1)+(0)+(1)+(2)=2.$ Sign interpret karna: jahan $f(x)<0$ (yaani $x<2$ ke liye) "height" negative hai, toh uska rectangle negative area contribute karta hai. Yeh Signed area hai: axis ke neeche wala area minus count hota hai. Yahan negative chunk $(-1)$ aur positive chunk $(+3)$ net karke $+2$ dete hain. Sach wala integral ${\int }_0^4(x-2)dx=\frac{x^2}{2}-2x{|}_0^4=8-8=0$ hai; hamara coarse $R_4=2$ overshoot karta hai (increasing function ⇒ right overestimates) lekin concept — signed contributions — wahi main point hai. ✓

Key idea: parent note ka "left over/under, right over/under" rule monotonic function ke liye hi kaam karta hai. $[0,\pi ]$ pe, $\cos x$ rise karta hai? Nahi — yeh $1$ se $-1$ tak girta hai. Yeh poore interval pe monotonic decreasing hai.

• Decreasing ⇒ left edge har strip ka sabse lamba point hai ⇒ Left overestimates.
• Decreasing ⇒ right edge sabse chhota hai ⇒ Right underestimates.
• Midpoint: cancellation ki wajah se sabse best. Exactly confirm karo: ${\int }_0^{\pi }\cos xdx=\sin x{|}_0^{\pi }=\sin \pi -\sin 0=0-0=0.$ Toh sach wali value $0$ hai; $L_n>0>R_n$ uske around hain. ✓ (Yeh Fundamental Theorem of Calculus ko ek oracle ki tarah use karta hai — checking ke liye legal hai, L5 ke from-scratch derivations ke liye nahi.)

Hum kya kar rahe hain: do sums ko term by term subtract karo aur telescope collapse hote dekho. $L_n={\sum }_{i=1}^{n}f(x_{i-1})\Delta x=\Delta x(f(x_0)+f(x_1)+\cdots +f(x_{n-1})),$ $R_n={\sum }_{i=1}^{n}f(x_i)\Delta x=\Delta x(f(x_1)+f(x_2)+\cdots +f(x_n)).$ Inhe line up kyon karein: har interior term $f(x_1),\dots ,f(x_{n-1})$ dono lists mein appear karta hai, toh difference mein cancel ho jaata hai. Sirf do ends bachte hain: $R_n-L_n=\Delta x(f(x_n)-f(x_0))=(f(b)-f(a))\Delta x.$ Matlab: $\Delta x=\frac{b-a}{n}\to 0$ jab $n\to \infty$, aur $f(b)-f(a)$ ek fixed number hai. Toh $R_n-L_n\to 0$: left aur right sums squeeze hokar ek hi limit ki taraf jaate hain — ek pehli jhalak ki integral well-defined kyon hai (dekho Continuity and Integrability). ✓

Strategy: pattern $\sum f(x_i)\Delta x$ se $f$, $a$, $b$, $\Delta x$ ke pieces reverse-engineer karo.

• Akela $\frac{1}{n}$ width $\Delta x$ hai. Kyunki $\Delta x=\frac{b-a}{n}$, iska matlab $b-a=1$ hai.
• Sample point $\frac{i}{n}$ hai. $a=0$ ke saath, $x_i=a+i\Delta x=\frac{i}{n}$ ✓ — $[0,1]$ pe ek right sum.
• Height $\sqrt{1+x}$ hai jo $x=\frac{i}{n}$ pe evaluate ki gayi hai, toh $f(x)=\sqrt{1+x}$. Isliye ${\lim }_{n\to \infty }{\sum }_{i=1}^n\frac{1}{n}\sqrt{1+\frac{i}{n}}={\int }_0^1\sqrt{1+x}dx.$ Evaluate karo ($(1+x{)}^{1/2}$ ka antiderivative $\frac{2}{3}(1+x{)}^{3/2}$ hai): ${\int }_0^1\sqrt{1+x}dx=\frac{2}{3}(1+x{)}^{3/2}{|}_0^1=\frac{2}{3}(2^{3/2}-1)=\frac{2}{3}(2\sqrt{2}-1)\approx \mathrm{1.2190.}$ Answer: $\frac{2}{3}(2\sqrt{2}-1)\approx 1.219$. Yeh exactly wahi trick hai jo Area under a curve problems reverse mein use karti hain. ✓

(a) Trapezoidal Rule left aur right ka average karta hai: $T_4=\frac{1}{2}(1.75+3.75)=\frac{1}{2}(5.5)=\mathrm{2.75.}$ (b) Errors (signed, sach − estimate): $\text{error}(T_4)=2.6667-2.75=-0.0833,\text{error}(M_4)=2.6667-2.625=+\mathrm{0.0417.}$ Sizes: $|-0.0833|=0.0833$ vs $0.0417$. Toh midpoint ki error trapezoid ki roughly half hai, aur opposite sign mein — yeh classic $2:1$ relationship hai. Answer: $T_4=2.75$; midpoint roughly factor $2$ se jeetta hai. ✓ (Zyada $n$ push karo aur dono shrink honge; dekho Simpson's Rule jo cleverly $M$ aur $T$ ko combine karta hai aur aur bhi zyada error khatam karta hai.)

Step 1 — grid set up karo. $a=1,b=3$, toh $\Delta x=\frac{3-1}{n}=\frac{2}{n}$, aur $x_i=a+i\Delta x=1+\frac{2i}{n}.$ Kyon: right sums inhi grid points pe evaluate karte hain.

Step 2 — sum likho. $f(x)=2x+1$ ke saath, $f(x_i)=2(1+\frac{2i}{n})+1=3+\frac{4i}{n}.$ $R_n={\sum }_{i=1}^n(3+\frac{4i}{n})\frac{2}{n}=\frac{2}{n}{\sum }_{i=1}^{n}3+\frac{2}{n}\cdot \frac{4}{n}{\sum }_{i=1}^{n}i.$ Hum aisa kyon split kar sakte hain: do cheezein add hoti hain uska sum har piece ka alag sum hota hai (addition regroup ho sakti hai), aur koi bhi factor jo counter $i$ pe depend nahi karta — yahan $\frac{2}{n}$ aur $\frac{4}{n}$ — woh har term mein same hai, toh $\sum$ ke bahar ek common multiplier ki tarah nikal sakta hai. Andar jo bacha woh ek pure $\sum i$ hai jiske liye hamare paas formula hai.

Step 3 — summation formulas apply karo. ${\sum }_{i=1}^{n}3=3n$ (constant $3$ ko $n$ baar add karna) aur ${\sum }_{i=1}^{n}i=\frac{n(n+1)}{2}$: $R_n=\frac{2}{n}(3n)+\frac{8}{n^2}\cdot \frac{n(n+1)}{2}=6+\frac{4(n+1)}{n}=6+4+\frac{4}{n}=10+\frac{4}{n}.$ Kyon: closed forms substitute karne se sum ordinary algebra mein badal jaata hai jise hum simplify kar sakte hain.

Step 4 — limit lo. Jab $n\to \infty$, tail $\frac{4}{n}\to 0$ (ek fixed top ek exploding bottom pe shrink hokar kuch nahi hota): ${\int }_1^3(2x+1)dx={\lim }_{n\to \infty }(10+\frac{4}{n})=10.$ Antiderivative se cross-check: $x^2+x{|}_1^3=(9+3)-(1+1)=12-2=10.$ ✓ Endpoints dhyaan rakho: answer difference $12-2=10$ hai, sirf upper value $12$ nahi — Riemann limit already tumhare liye lower endpoint subtract kar deta hai. ✓

Step 1 — grid. $\Delta x=\frac{1}{n}$, $x_i=\frac{i}{n}$. Step 2 — sum. $R_n={\sum }_{i=1}^n(\frac{i}{n}{)}^3\frac{1}{n}=\frac{1}{n^4}{\sum }_{i=1}^n{i}^3.$ $\frac{1}{n^4}$ front pe kyon aata hai: $\frac{1}{n}$ ko cube karne se $\frac{1}{n^3}$, times $\Delta x$ se $\frac{1}{n}$, aur iska koi bhi part counter $i$ pe depend nahi karta, toh yeh $\sum$ ke bahar nikal jaata hai aur pure $\sum i^3$ bach jaata hai. Step 3 — $\sum i^3={\left[\frac{n(n+1)}{2}\right]}^2=\frac{n^2(n+1{)}^2}{4}$ plug karo: $R_n=\frac{1}{n^4}\cdot \frac{n^2(n+1{)}^2}{4}=\frac{(n+1{)}^2}{4n^2}=\frac{n^2+2n+1}{4n^2}.$ Step 4 — limit. Top aur bottom ko $n^2$ se divide karo (sabse tezi se badhne wala power, toh bachne wale pieces constants ban jaate hain): $\frac{1+2/n+1/n^2}{4}\to \frac{1}{4}$. ${\int }_0^1{x}^{3}dx=\frac{1}{4}.✓$

(a) Zero-width interval — mesh ke through argue kiya. Hum $\Delta x\equiv 0$ set karke partition nahi keh sakte; definition genuine partitions ke saath kaam karti hai jinka mesh $\Vert P\Vert ={\max }_i\Delta x_i$ phir $0$ ki taraf drive kiya jaata hai. Toh interval $[a,a]$ ka koi bhi honest partition lo. Lekin $[a,a]$ ek single point hai: uski total length $b-a=a-a=0$ hai, aur uske subintervals ki widths non-negative numbers hain jo sum karke $0$ deti hain, isliye har $\Delta x_i=0$ hai. Isliye, kisi bhi partition ke liye (uska mesh already $0$ hai) aur kisi bhi sample points $x_i^{\ast }$ ke choice ke liye, $S_n={\sum }_{i=1}^{n}f(x_i^{\ast })\Delta x_i={\sum }_{i=1}^{n}f(x_i^{\ast })\cdot 0=0.$ Har admissible Riemann sum $0$ ke barabar hai, toh $\Vert P\Vert \to 0$ ke saath limit trivially $0$ hai: ${\int }_a^{a}f(x)dx=0.$ Point: humne assume nahi kiya "$\Delta x=0$ ek partition hai." Humne ek degenerate interval ke legitimate partitions dekhe aur paaya ki widths interval ki zero length ki wajah se $0$ hone pe majboor hain. ✓

(b) Dirichlet function. Lo $D(x)=\{\begin{array}{ll}1,& x\text{ rational}\\ 0,& x\text{ irrational.}\end{array}$ Yeh bounded hai ($0\le D\le 1$). $[0,1]$ ka koi bhi partition fix karo $n$ strips mein width $\Delta x=\frac{1}{n}$ ke saath. Har subinterval mein, chahe kitna bhi tiny ho, dono ek rational aur ek irrational number hain (dono kinds dense hain — dekho Limits of sequences). Definition demand karti hai ki sample points $x_i^{\ast }$ ki har legal choice ke liye same limit ho. Do legal choices test karo:

• Saare $x_i^{\ast }$ rational: har height $D(x_i^{\ast })=1$, toh $S_n={\sum }_{i=1}^{n}1\cdot \Delta x=n\cdot \frac{1-0}{n}=1\text{har }n\text{ ke liye}\Rightarrow {\lim }_{\Vert P\Vert \to 0}{S}_n=1.$
• Saare $x_i^{\ast }$ irrational: har height $D(x_i^{\ast })=0$, toh $S_n={\sum }_{i=1}^{n}0\cdot \Delta x=0\text{har }n\text{ ke liye}\Rightarrow {\lim }_{\Vert P\Vert \to 0}{S}_n=0.$ Do legal sampling choices limits $1$ aur $0$ deti hain. Kyunki $1\ne 0$, koi single number $I$ nahi hai jo saari choices ke liye limit ho, toh $D$ integrability requirement fail karta hai — yeh Riemann integrable nahi hai. Yahi exactly woh reason hai ki parent ka phrase "sample points ki har choice ke liye" definition ka asli dum hai, decoration nahi (compare karo Continuity and Integrability se, jahan continuity ek common limit ki guarantee karti hai). ✓