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

4.2.3 · D3 · Maths › Calculus II — Integration › Riemann sums — left, right, midpoint; formal definition of d

Yeh Riemann sums & the formal definite integral ka ek Deep Dive child hai. Parent note ne machinery banayi thi (partitions, sample points, limit). Yahan hum use stress-test karte hain har us tarah ke input ke against jo is topic mein aa sakta hai — positive area, negative area, mixed signs, ek degenerate zero-width case, ek limit jo scratch se compute ki gayi, ek word problem with units, aur ek exam twist. Steps padhne se pehle guess karo.

The scenario matrix

Kuch bhi work karne se pehle, yeh ek map hai har case class ka jo is topic mein aa sakti hai. Neeche har worked example us cell(s) ke saath tagged hai jo woh cover karta hai. Agar tum har row handle kar sako, toh exam mein koi cheez surprise nahi kar sakti.

Do ideas baar baar aate hain, toh pehle unhe plain words mein nail karte hain:

Example 1 — Positive function, from the definition (cell A)

Forecast: parent ne ${\int }_0^1{x}^2=\frac{1}{3}$ nikala tha. $[0,2]$ tak stretch karna $x^3$ ko scale karta hai, toh kuch $\frac{8}{3}\approx 2.67$ ke paas guess karo. Yeh yaad rakho.

1. Mesh set up karo. $\Delta x=\frac{b-a}{n}=\frac{2-0}{n}=\frac{2}{n}$, aur $x_i=a+i\Delta x=\frac{2i}{n}$. Yeh step kyun? Right rule right edge $x_i$ sample karta hai; hume isko sum karne se pehle $i$ aur $n$ mein formula ke roop mein chahiye.
2. Sum likho. $R_n={\sum }_{i=1}^{n}f(x_i)\Delta x={\sum }_{i=1}^n{\left(\frac{2i}{n}\right)}^2\frac{2}{n}=\frac{8}{n^3}{\sum }_{i=1}^n{i}^2.$ Yeh step kyun? Har constant ($2^2\cdot 2=8$, aur $\frac{1}{n^2}\cdot \frac{1}{n}=\frac{1}{n^3}$) bahar nikalo, ek clean sum of squares bachta hai.
3. Closed form insert karo ${\sum }_{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}$: $R_n=\frac{8}{n^3}\cdot \frac{n(n+1)(2n+1)}{6}=\frac{8(n+1)(2n+1)}{6n^2}=\frac{4(2n^2+3n+1)}{3n^2}.$ Yeh step kyun? Ek finite sum jiska limit nahi le sakte; ek rational function of $n$ jiska le sakte hain.
4. Limit lo $n\to \infty$ (top aur bottom ko $n^2$ se divide karo): ${\int }_0^2{x}^{2}dx={\lim }_{n\to \infty }\frac{4(2+\frac{3}{n}+\frac{1}{n^2})}{3}=\frac{4\cdot 2}{3}=\frac{8}{3}.$ Yeh step kyun? Strips ko infinitely thin banana ($n\to \infty$) integral ki definition hai; $\frac{3}{n},\frac{1}{n^2}$ terms vanish ho jaate hain.

Verify: Fundamental Theorem of Calculus deta hai $\frac{x^3}{3}{|}_0^2=\frac{8}{3}\approx 2.667$. Forecast se match karta hai. ✓

Example 2 — Increasing $f$: Left under, Right over (cell B)

Forecast: $x^2$ $[0,2]$ par rise karta hai, toh Left undershoot karna chahiye, Right overshoot.

1. Mesh. $\Delta x=\frac{2}{4}=0.5$; nodes $x_0..x_4=0,0.5,1,1.5,2$. Kyun? Char equal strips; dono rules ke liye shared edge list chahiye.
2. Left $x_0,x_1,x_2,x_3=0,0.5,1,1.5$ sample karta hai: $L_4=0.5(0^2+0.5^2+1^2+1.5^2)=0.5(0+0.25+1+2.25)=0.5(3.5)=\mathrm{1.75.}$ Kyun? Left rule har strip ka left edge $x_{i-1}$ use karta hai — rising curve ka chhota side.
3. Right $x_1,x_2,x_3,x_4=0.5,1,1.5,2$ sample karta hai: $R_4=0.5(0.5^2+1^2+1.5^2+2^2)=0.5(0.25+1+2.25+4)=0.5(7.5)=\mathrm{3.75.}$ Kyun? Right rule right edge $x_i$ use karta hai — rising curve ka lamba side.

Verify: $L_4=1.75<2.667<3.75=R_4$. Bilkul predicted straddle. Yeh bhi note karo ki $R_4-L_4=0.5(f(2)-f(0))=0.5\cdot 4=2$ — total error gap equals width $\times$ height-jump, ek handy check. ✓

Example 3 — Decreasing $f$: ordering ulta ho jaata hai (cell C)

Forecast: $\frac{1}{x}$ fall karta hai, toh ab Left tall side hai (over) aur Right short side (under) — Ex 2 ka mirror.

1. Mesh. $\Delta x=\frac{3-1}{2}=1$; nodes $1,2,3$; midpoints $1.5,2.5$. Kyun? Do strips; edges aur centres chahiye.
2. Left ($x=1,2$): $L_2=1\cdot (f(1)+f(2))=1+0.5=1.5$. Kyun? Falling curve ke left edges true height se upar baithe hain ⇒ overestimate.
3. Right ($x=2,3$): $R_2=1\cdot (f(2)+f(3))=0.5+\frac{1}{3}=\frac{5}{6}\approx 0.8333$. Kyun? Falling curve ke right edges neeche baithe hain ⇒ underestimate.
4. Midpoint ($x=1.5,2.5$): $M_2=1\cdot (f(1.5)+f(2.5))=\frac{2}{3}+\frac{2}{5}=\frac{16}{15}\approx 1.0667$. Kyun? Centre sample karne se over-hang aur under-hang roughly cancel ho jaate hain.

Verify: $1.5>1.0986>0.8333$ ✓, aur $|1.0667-1.0986|=0.0319$ dono edge errors se kaafi chhota hai — midpoint jeet jaata hai, jaise promise tha. ✓

Example 4 — Negative function ⇒ negative integral (cell D)

Forecast: curve $y=-x^2$ poori tarah axis ke neeche hai, toh iska "area" negative aana chahiye — precisely $-\frac{8}{3}$.

1. Ex 1 ka skeleton reuse karo. $\Delta x=\frac{2}{n}$, $x_i=\frac{2i}{n}$, lekin ab $f(x_i)=-{\left(\frac{2i}{n}\right)}^2$. Kyun? Sirf ek global minus sign har height par hai; widths positive rehti hain.
2. Sum. $R_n={\sum }_{i=1}^n-{\left(\frac{2i}{n}\right)}^2\frac{2}{n}=-\frac{8}{n^3}{\sum }_{i=1}^n{i}^2=-\frac{4(2n^2+3n+1)}{3n^2}.$ Kyun? Linearity: $-1$ bahar nikalna sirf Ex 1 result ko negate karta hai.
3. Limit. ${\int }_0^2-x^{2}dx=-\frac{8}{3}$. Kyun? Ex 1 jaisi hi limit sign carry karke.

Verify: ek region ka signed area jo poori tarah axis ke neeche hai zaroori negative hona chahiye; magnitude $\frac{8}{3}$ Ex 1 ke positive twin se match karta hai. FTC: $-\frac{x^3}{3}{|}_0^2=-\frac{8}{3}$. ✓ Figure dekho: poora shaded region white axis ke neeche baitha hai, toh har rectangle ek minus sign carry karta hai — isliye total $-\frac{8}{3}$ hai, $+\frac{8}{3}$ nahi.

Example 5 — Interval ke andar sign change (cell E)

Forecast: $x-1$ $[0,1]$ par negative hai aur $[1,2]$ par positive, equal size ke mirror-image triangles ke through. Guess karo dono signed areas cancel hokar $0$ denge.

1. Do regions identify karo. $[0,1]$ par below-axis triangle jiske legs $1$ aur $1$ hain; $[1,2]$ par above-axis triangle jiske legs $1$ aur $1$ hain. Kyun? $x-1=0$ at $x=1$; woh root positive aur negative area ko alag karta hai.
2. Har piece ka signed area. Below piece: $-\frac{1}{2}(1)(1)=-\frac{1}{2}$. Above piece: $+\frac{1}{2}(1)(1)=+\frac{1}{2}$. Kyun? Right triangle ka area $\frac{1}{2}\text{base}\times \text{height}$ hota hai; left wala minus carry karta hai kyunki wahan $f<0$ hai.
3. Add karo. ${\int }_0^2(x-1)dx=-\frac{1}{2}+\frac{1}{2}=0$. Kyun? Integral signed pieces ka algebraic total hai, total geometric area nahi.

Verify: FTC $\frac{x^2}{2}-x{|}_0^2=(2-2)-(0)=0$. ✓ Dhyan raho: total geometric area (agar paint chahiye ho) $\frac{1}{2}+\frac{1}{2}=1\ne 0$ hai — ek classic trap. Figure dekho: red triangle (left, below axis) aur green triangle (right, above axis) congruent hain, toh unke signed contributions $-\frac{1}{2}$ aur $+\frac{1}{2}$ annihilate ho jaate hain.

Example 6 — Degenerate zero-width interval (cell F)

Forecast: koi width nahi matlab koi rectangle nahi matlab koi area nahi. Guess $0$.

1. Mesh dekho. $a=b=3$ ke saath: $\Delta x=\frac{b-a}{n}=\frac{0}{n}=0$ har $n$ ke liye. Kyun? Interval ki length $0$ hai; har subinterval ek point par collapse ho jaata hai.
2. Har term khatam ho jaata hai. $S_n={\sum }_{i=1}^{n}f(x_i^{\ast })\cdot 0=0$ heights ki parwah kiye bina. Kyun? Height × zero width = zero, chahe $f$ kitni bhi tall ho.
3. Zeros ki limit zero hai. ${\int }_3^{3}f=\underset{n\to \infty }{\lim }0=0$. Kyun? Ek constant sequence us constant par converge karti hai.

Verify: FTC se consistent, $F(3)-F(3)=0$ kisi bhi antiderivative $F$ ke liye. Yeh degenerate rule hi hai jo ${\int }_a^b={\int }_a^c+{\int }_c^b$ ko tab bhi hold karwata hai jab $c=a$ ho. ✓

Example 7 — Midpoint straight line ke liye exact hai (cell G)

Forecast: midpoint lines ke liye exact hota hai, toh ek strip kaafi honi chahiye. Exact answer guess karo.

1. Ek strip. $\Delta x=2$; single subinterval $[0,2]$ hai; midpoint $=1$. Kyun? $n=1$ ⇒ poora interval ek strip hai; uska centre $\frac{0+2}{2}=1$ hai.
2. Evaluate karo. $M_1=f(1)\cdot 2=(3\cdot 1+1)\cdot 2=4\cdot 2=8$. Kyun? Riemann term = height at midpoint × width.

Verify: FTC $\frac{3x^2}{2}+x{|}_0^2=(6+2)-0=8$. ✓ Exact kyun: ek straight line par woh triangle jo midpoint rectangle ke upar nikalta hai exactly us triangle ke barabar hai jo neeche miss hota hai — perfect cancellation, kisi bhi $n$ ke liye. Figure dekho.

Example 8 — Word problem with units (cell H)

Forecast: distance = speed-vs-time graph ke neeche area. Speed badhti hai, toh right sum true distance ko thoda overshoot karna chahiye.

1. Integral kyun? Distance = accumulated speed $={\int }_0^{3}v(t)dt$; har rectangle $\text{(m/s)}\times \text{(s)}=\text{m}$ hai, toh sum metres carry karta hai. Kyun? Units audit: height $\times$ width woh quantity dena chahiye jo hum chahte hain.
2. Right sum. $\Delta t=1$ s; right nodes $t=1,2,3$: $R_3=1\cdot (v(1)+v(2)+v(3))=24+28+32=84\text{m}.$ Kyun? Right rule har 1-second strip ka right edge sample karta hai.
3. Exact. ${\int }_0^3(20+4t)dt=[20t+2t^2{]}_0^3=60+18=78\text{m}.$ Kyun? $20+4t$ ka antiderivative $20t+2t^2$ hai; ends par evaluate karo.

Verify: $R_3=84>78$ — overshoot, jaisa increasing speed ke liye forecast tha. Units: poore mein metres. ✓ (Yahan midpoint sum exact $78$ dega kyunki $v$ linear hai — phir cell G.)

Example 9 — Exam twist: ek limit ko integral ke roop mein padhna (cell I)

Forecast: yeh ek right Riemann sum jaisa lagta hai jisme $\Delta x=\frac{1}{n}$ length $1$ ke interval par hai. Ise decode karo, sum mat karo.

1. Template match karo ${\int }_a^{b}f=\underset{n\to \infty }{\lim }\sum f(a+i\Delta x)\Delta x$. Yahan $\Delta x=\frac{1}{n}$, toh $b-a=1$; $a=0,b=1$ lo. Kyun? $\Delta x=\frac{b-a}{n}$; $\frac{1}{n}$ match karna interval width ko $1$ force karta hai.
2. $f$ padhao. Sample point $a+i\Delta x=\frac{i}{n}$ hai, aur height $\sqrt{1+\frac{i}{n}}=\sqrt{1+x}$ hai. Toh $f(x)=\sqrt{1+x}$ on $[0,1]$. Kyun? $\frac{i}{n}$ ko continuous variable $x$ se replace karo; yahi limit exactly karta hai.
3. Integral evaluate karo. ${\int }_0^1\sqrt{1+x}dx=[\frac{2}{3}(1+x{)}^{3/2}{]}_0^1=\frac{2}{3}(2^{3/2}-1)=\frac{2}{3}(2\sqrt{2}-1).$ Kyun? $(1+x{)}^{1/2}$ ka antiderivative $\frac{2}{3}(1+x{)}^{3/2}$ hai; ends par evaluate karo.

Verify: numerically $\frac{2}{3}(2\sqrt{2}-1)\approx \frac{2}{3}(1.8284)\approx 1.2190$; large $n$ ke saath ek coarse numeric sum isko confirm karta hai. ✓ Yeh "backwards" reading ek favourite exam device hai.

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