Riemann Sums: Area ka Approximation
Transcript
Aaj hum Riemann sums ko deep dive mein dekhenge: curve ke neeche ka area, rectangles ki finite sum se kaise nikalte hain. Pehle ek curve dekhte hain, f of x equal x squared by two, interval zero se do tak. Shaded region hamara target hai. Sawaal yeh hai: iska exact area kaise define karein, jab boundary curved ho?
\text{Area} = ?Idea simple hai: region ko partition karo, har piece mein rectangle lagao, aur unke areas ko add karo. Interval a se b ko n subintervals mein todte hain. Yahan a zero hai, b do hai, aur n chaar hai. Equal width ke liye delta x equal b minus a over n. Humare case mein do minus zero, bata chaar, yaani half.
- Partition, rectangle, add
\Delta x = \frac{b-a}{n} = \frac{2-0}{4} = \frac{1}{2}Har subinterval mein ek sample point x star i chunte hain. Rectangle ki height hogi f of x star i.
\text{height} = f(x_i^*),\quad \text{width} = \Delta xEk rectangle ka area height times width hai. Saare rectangles ko add karo, wahi Riemann sum hai.
S_n = \sum_{i=1}^{n} f(x_i^*)\,\Delta xSample point ke choices hain: left endpoint, right endpoint, ya midpoint. Teeno alag sums dete hain, par ek hi integral ki taraf jaate hain. Ab worked example: left Riemann sum, n equal chaar. Left endpoints hain zero, half, one, aur one point five.
- Left, Right, Midpoint
L_4 = \tfrac12\left[f(0)+f(0.5)+f(1)+f(1.5)\right]Function values nikalte hain: f of zero zero, f of half one-eighth, f of one half, f of one point five, nine-eighths.
L_4 = \tfrac12\left[0+\tfrac18+\tfrac12+\tfrac98\right]Bracket ke andar one-eighth plus four-eighths plus nine-eighths, yaani fourteen-eighths, matlab seven-fourths.
L_4 = \tfrac12\cdot\tfrac74Toh left sum seven-eighths hai, yaani zero point eight seven five. Yeh underestimate hai kyunki function increasing hai.
L_4 = \tfrac78 = 0.875Right sum ke liye endpoints half, one, one point five, two lo. Values one-eighth, half, nine-eighths, aur do.
R_4 = \tfrac12\left[\tfrac18+\tfrac48+\tfrac98+\tfrac{16}{8}\right] = \tfrac12\cdot\tfrac{30}{8} = \tfrac{15}{8}=1.875Exact area four-thirds hai, yaani one point three three three. Left chhota, right bada, true value dono ke beech.
0.875 < \tfrac43 \approx 1.333 < 1.875Ab why: increasing function par left rectangle curve ke neeche rehta hai, right rectangle upar nikal jaata hai. Isliye bound milta hai. Jab n badhta hai, delta x chhota hota hai aur rectangles curve ko behtar fit karte hain. Error shrink hota hai. Limit lene par definite integral milta hai. Yeh Riemann integral ki definition hai, jab limit exist kare.
\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^*)\,\Delta xEdge case: agar function decreasing ho, toh left sum overestimate karta hai aur right sum underestimate. Sirf monotonic functions par yeh rule lagta hai. Common misconception: log sochte hain ki left sum hamesha underestimate hai. Yeh galat hai, yeh sirf increasing function par sahi hai. Real-life mein velocity graph ke neeche ka area displacement deta hai, aur rectangles discrete speed readings ke sums hain.
- Decreasing: left over, right under
Exam tip: free response mein pehle sum ka setup likho, delta x aur har term, kyunki sirf final number se points jaa sakte hain. Ab ek AP-style sawaal, jisme dhyan se sochna hai ki sum overestimate hai ya underestimate. Summary: Riemann sum rectangles ke areas ka sum hai, aur n infinity jaane par limit definite integral ban jaati hai.
- Exam tip: setup likho
Quick quiz
Q1. f is positive and decreasing on [a, b]. A right Riemann sum with n equal subintervals approximates the integral of f. Which statement is true?
Misconception: right sums always overestimate. For a decreasing function the right endpoint gives the smallest height, so it underestimates.
