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

True — har strip mein left edge ek rising curve ka sabse nichla point hota hai, isliye har rectangle curve ke neeche baitha rehta hai. Jaise hi function badhna band kar deti hai ye fail ho jaata hai, isliye "increasing" zaroori hai.

True — ek line par jo triangle midpoint height ke upar bahar nikal jaata hai woh exactly us triangle se match karta hai jo neeche miss hota hai, isliye errors har strip par cancel ho jaate hain, chahe kitni bhi strips use karo.

Dhyan se — yeh akela sufficient nahi hai; definition maangti hai ki limit har sample choice ke liye same ho, sirf left aur right ke liye nahi. Practice mein continuous $f$ ke liye yeh kaafi hai, lekin ek logical statement ke roop mein yeh false hai (ek function edges par agree kar sakta hai lekin clever samples ke liye diverge kar sakta hai).

False — heights akeli area nahi hoti. Tumhe har height ko strip ki width se multiply karna hoga: area hai $\sum f(x_i^{\ast })\Delta x_i$. $\Delta x$ drop karne se ek aisa number milta hai jiske units bilkul galat hain.

Generally False — non-uniform partition ke liye tum bahut saari choti strips add kar sakte ho jabki ek moti strip ko untouched chod do, isliye $\Vert P\Vert$ kabhi nahi shrink hoti. Sirf uniform partitions ke liye hi $n\to \infty$ se $\Vert P\Vert \to 0$ hota hai.

True — yeh signed area compute karta hai, isliye $f$ ka jo bhi hissa $x$-axis ke neeche hai woh negatively contribute karta hai. Signed area dekho; integral area tabhi hota hai jab $f\ge 0$ on $[a,b]$.

True — yeh ek core theorem hai (dekho Continuity and Integrability). Lekin converse false hai: ek function integrable ho sakti hai jabki usmein jumps hoon, isliye continuity sufficient hai, necessary nahi.

False — woh average Trapezoidal rule hai, midpoint rule nahi (dekho Trapezoidal Rule). Ye genuinely alag approximations hain, aur midpoint aksar trapezoid ko beat karta hai halanki har strip mein sirf ek sample use karta hai.

False — uniform partition ke liye $R_n-L_n=\Delta x(f(b)-f(a))$, isliye sums tab equal hote hain jab $f(a)=f(b)$, jo bahut si non-constant functions satisfy karti hain (jaise ek symmetric bump).

True — monotone function ke liye left over/under-estimate karta hai aur right ulta karta hai, isliye sach in dono ke beech mein hota hai; midpoint, sach ke paas hone ki wajah se, in dono ke beech mein bhi baithta hai. (Non-monotone $f$ ke liye yeh ordering toot sakti hai.)

Strip $i$ ka left edge $x_{i-1}=a+(i-1)\Delta x$ hai, $x_i$ nahi. $x_i$ use karna silently right sum compute kar deta hai — ek classic off-by-one.

Tum easy samples tabhi pick kar sakte ho jab yeh jaanta ho ki $f$ integrable hai, kyunki integrability guarantee karti hai ki sabhi choices same limit deti hain. Integrability prove karne ke liye khud samples choose karna circular hai — tumhe pehle pata hona chahiye ki limit choice-independent hai.

$x^2$ $[0,1]$ par increasing hai, isliye right sum overestimate karta hai — har $R_n>1/3$ hai, upar se approach karta hua. Right sums sirf decreasing functions ke liye undercount karte hain.

Yeh Riemann integrable hi nahi hai: rationals sample karne par sum $1$ milta hai, irrationals sample karne par $0$ milta hai, isliye koi single limit exist nahi karta. "Almost everywhere" ek Lebesgue idea hai, Riemann ka nahi.

Zyada samples ka matlab zyada accurate nahi hota. Midpoint ka single centred sample symmetric error cancellation se benefit karta hai; uski error typically trapezoid ki aadhi hoti hai aur opposite sign ki. Count of samples ≠ quality.

Yeh vanish nahi hota — yeh $dx$ ban jaata hai. $\int fdx$ mein $dx$ symbol $\Delta x$ width ka surviving trace hai limit ke baad, isliye ise bhoolna itni serious error hai.

Continuity sufficient hai lekin necessary nahi. Finitely many jumps wali function (jaise ek step function) phir bhi integrable hai, kyunki misbehaving strips ka total width zero ho jaata hai jab $\Vert P\Vert \to 0$.

Kyunki agar left, right, aur koi bhi aur sampling sab agree karein, toh answer $f$ ki genuine property reflect karta hai na ki humari arbitrary choice ka artefact. Ek choice-dependent "answer" the area ke roop mein meaningless hoga.

Average small reh sakta hai jabki ek stubborn moti strip wahan approximation bigaad deti hai. Maximum ko zero force karna guarantee karta hai ki har strip vanish ho, isliye staircase curve ko har jagah hug karti hai.

Yeh Sum ke liye stand karta hai — integral Riemann sum $\sum f(x_i^{\ast })\Delta x$ ki limit hai, isliye Leibniz ne summation ko ek stretched S mein badal diya aur discrete $\Delta x$ ko infinitesimal $dx$ mein.

Centred sample strip ke curve ke ek side ke upar ke overshoot ko doosri side ke undershoot se cancel karne deta hai. Left/right samples ek extreme edge par baithe hote hain, isliye unki errors ek direction mein accumulate hoti hain bina kisi cancellation ke.

Koi bhi finite $n$ staircase aur curve ke beech ek nonzero gap chodta hai, isliye yeh approximation hai, exact area nahi. Sirf limit $n\to \infty$ hi error ki aakhri sliver remove karta hai; wahi exact value hai jo "definite integral" ka matlab hai.

Ek line par midpoint height strip ke paas average height ke barabar hoti hai, isliye rectangle area sahi (trapezoidal) area se exactly equal hoti hai. Left aur right heights average nahi hoti, isliye woh slope times half the width se over/undershoot karte hain.

Yeh $0$ hoga — interval ki zero width hai, isliye har $\Delta x_i=0$ hai aur har rectangle ka area zero hai. Measure karne ke liye literally koi region hi nahi hai.

Integral $0$ hai: jaise strips pateeli hoti hain, kisi bhi sample ka exact spike par land karne ki probability vanish ho jaati hai, aur ek strip jo usse contain bhi kare woh height $\times$ (zero ki taraf shrink hoti width) contribute karti hai. Ek single point ka koi area nahi hota.

Teeno identical hain aur exactly $c(b-a)$ ke barabar hain har $n$ ke liye, kyunki height $c$ hai chahe tum kahan bhi sample karo. Ek flat function mein koi rise nahi hai jise koi bhi rule miss kare.

Jahan $f<0$ wahan strips negative rectangle areas contribute karti hain, isliye sum (aur limit) signed area compute karta hai — positive parts minus negative parts, total geometric area nahi. Dekho Area under a curve vs Signed area.

Nahi — fat strip error ko dominate karti hai aur $\Vert P\Vert$ bada hai, isliye wahan approximation kharab hai. Accuracy sabse wide strip se control hoti hai, na ki kitni strips hain us se.

Yeh $0$ hai. Kyunki $L_n-R_n=\Delta x(f(a)-f(b))$ aur $\Delta x\to 0$ while $f(a),f(b)$ fixed aur finite hain, poora difference collapse ho jaata hai — dono sums same integral par converge karte hain.