Average value of a function: Worked examples — Hinglish deep dive

4.2.18 · D3 · Maths › Calculus II — Integration › Average value of a function

Scenario matrix

Har "average value" problem in cells mein se kisi ek mein aati hai. Hum sabhi ko cover karenge.

#	Case class	Isme kya tricky hai	Example jo ise cover karta hai
C1	$f \geq 0$ everywhere	baseline, area honest area hai	Ex 1
C2	$f \leq 0$ everywhere	integral negative hai → average negative	Ex 2
C3	$f$ ka sign change hota hai, symmetric	positive & negative bits cancel hote hain → average $= 0$	Ex 3
C4	$f$ ka sign change hota hai, not symmetric	partial cancellation → chota nonzero average	Ex 4
C5	interval negative $x$ mein	limits jaise $[- 2, - 1]$ , width dhyan se dekho	Ex 5
C6	degenerate: $b - a \to 0$	formula $0/0$ ban jaata hai → limiting value	Ex 6
C7	real-world word problem	tumhe khud $a, b, f$ dhundhne padenge	Ex 7
C8	exam twist: $c$ dhundho (MVT)	$f (c) = f_{avg}$ solve karo, root ko range mein rakho	Ex 8

Last cell Mean Value Theorem for Integrals (yahan se MVT abbreviated — yeh woh theorem hai jo guarantee karta hai ki ek special interior point $c$ hoga jahan function apne average ke barabar hoga) use karta hai.

Shuru karne se pehle, ek cheez dil mein basa lo: integral ek signed area hai. Curve ke jo bits $x$ -axis ke upar hain woh positive area count hote hain; jo bits neeche hain woh negative area count hote hain. Average value woh signed area hai jo width se divide hoti hai. Neeche diya figure exactly yahi dikhata hai.

Figure 1 — Ek wiggly $f$ jo $x$ -axis ko cross karti hai. Green shading woh piece hai jo axis ke upar hai (yeh $+$ count hoti hai); red shading woh piece hai jo neeche hai (yeh $-$ count hoti hai). Integral hai green-area minus red-area, aur average hai woh signed total jo width ke across spread hota hai. Yeh ek picture is page ke har case ka engine hai.

C1 — $f \geq 0$ everywhere (honest baseline)

Worked example Example 1 —

$f (x) = x^{2}$ on $[1, 4]$ Forecast: $x^{2}$ $1$ se $16$ tak jaata hai. Yeh ek aisi curve hai jo tezi se badhti hai, isliye yeh top ke paas zyada time spend karti hai. Guess karo average $1$ aur $16$ ke midpoint (jo $8.5$ hai) se kaafi upar hoga? Actually curves jo accelerate karti hain woh low pe zyada time spend karti hain, toh $8.5$ se thoda neeche guess karo. Yeh soch pakad ke rakho.

Step 1. Formula likho $a = 1, b = 4$ ke saath: $f_{avg} = \frac{1}{4 - 1} \int_{1}^{4} x^{2} d x .$ Yeh step kyun? Endpoints slot in karo taaki machinery ke paas numbers hon.

Step 2. $x^{2}$ ka antiderivative $\frac{x ^{3}}{3}$ hai (power rule, Fundamental Theorem of Calculus se): $\int_{1}^{4} x^{2} d x = [\frac{x ^{3}}{3}]_{1}^{4} = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21.$ Yeh step kyun? FTC kehta hai antiderivative ko top minus bottom pe evaluate karo.

Step 3. Width $4 - 1 = 3$ se divide karo: $f_{avg} = \frac{21}{3} = 7.$ Yeh step kyun? Integral $21$ ek area hai, average nahi; width $b - a$ se divide karne par woh total interval ke across evenly spread ho jaata hai aur average height milti hai. Answer: $f_{avg} = 7$ .

Verify: $7$ $min = 1$ aur $max = 16$ ke beech mein hai ✓, aur endpoint-midpoint $8.5$ se neeche hai — exactly forecast ki tarah, kyunki $x^{2}$ concave-up hai toh woh low values ke paas rehti hai. Units none (pure numbers). ✓

C2 — $f \leq 0$ everywhere (negative average)

Worked example Example 2 —

$f (x) = - 2 x$ on $[1, 3]$ Forecast: poori curve axis ke neeche baihthi hai ( $x > 0$ ke liye, $- 2 x < 0$ ). Signed area negative hai, toh average zaroor negative hoga. Yeh $- 2$ se $- 6$ tak ek straight line hai; line ke liye average midpoint value hoti hai, toh guess karo $- 4$ .

Step 1. Formula, $a = 1, b = 3$ : $f_{avg} = \frac{1}{3 - 1} \int_{1}^{3} (- 2 x) d x .$ Yeh step kyun? Same recipe — $f$ ka sign procedure nahi badalta, sirf result badalta hai.

Step 2. $- 2 x$ ka antiderivative $- x^{2}$ hai: $\int_{1}^{3} (- 2 x) d x = [- x^{2}]_{1}^{3} = (- 9) - (- 1) = - 8.$ Yeh step kyun? Yeh negative number hai hi signed area — region axis ke neeche rehta hai.

Step 3. Width $2$ se divide karo: $f_{avg} = \frac{- 8}{2} = - 4.$ Yeh step kyun? Same reason as always — integral $- 8$ ek signed area hai; width $b - a = 2$ se divide karne par woh average height mein convert hoti hai, aur kyunki area negative tha toh average bhi negative aata hai. Answer: $f_{avg} = - 4$ .

Verify: line-midpoint forecast se match karta hai: $x = 2$ pe, $f (2) = - 4$ ✓. Negative, jaisa axis ke neeche completely rehne wali curve ke liye zaroori tha. ✓

C3 — sign change with symmetry (average exactly zero)

Worked example Example 3 —

$f (x) = sin x$ on $[0, 2 π]$ Forecast: pehle half $[0, π]$ pe sine ek positive hump hai; doosre half $[π, 2 π]$ pe yeh ek identical negative hump hai. Dono areas mirror images hain. Guess karo average exactly $0$ hoga.

Figure 2 — $sin x$ over $[0, 2 π]$ . $[0, π]$ pe green lobe ka signed area $+ 2$ hai; $[π, 2 π]$ pe red lobe ka signed area $- 2$ hai. Yeh perfect mirror images hain, toh cancel ho jaate hain aur dashed orange line resulting average exactly $0$ mark karti hai.

Step 1. Formula, $a = 0, b = 2 π$ : $f_{avg} = \frac{1}{2 π} \int_{0}^{2 π} sin x d x .$ Yeh step kyun? Hum average-value formula ko diye gaye endpoints $a = 0$ , $b = 2 π$ ke saath likhte hain taaki abstract recipe ek concrete computation ban sake jo hum carry out kar sakein.

Step 2. $sin x$ ka antiderivative $- cos x$ hai: $\int_{0}^{2 π} sin x d x = [- cos x]_{0}^{2 π} = - cos (2 π) + cos 0 = - 1 + 1 = 0.$ Yeh step kyun? Positive hump ( $+ 2$ ) aur negative hump ( $- 2$ ) exactly cancel ho jaate hain — figure mein dono shaded lobes dekho.

Step 3. $2 π$ se divide karo: $f_{avg} = \frac{0}{2 π} = 0.$ Yeh step kyun? Signed area ko width $b - a = 2 π$ se divide karne par total average height mein convert hoti hai; kyunki area $0$ tha, average $0$ hai chahe interval kitna bhi wide ho. Answer: $f_{avg} = 0$ .

Verify: $0$ average ka matlab function $0$ hai nahi — iska matlab hai ups aur downs balance hain. Parent note ke $[0, π]$ case se compare karo jahan average $\frac{2}{π}$ tha; yahan humne mirror half add ki aur woh cancel ho gayi. ✓

C4 — sign change without symmetry (chota nonzero average)

Worked example Example 4 —

$f (x) = x^{3}$ on $[- 1, 2]$ Forecast: $[- 1, 0]$ pe cube negative hai; $[0, 2]$ pe yeh positive hai aur $8$ tak tezi se badhta hai. Positive part kaafi bada hai, toh average positive hoga lekin bahut bada nahi. Guess karo around $1$ .

Step 1. Formula, $a = - 1, b = 2$ , width $= 2 - (- 1) = 3$ : $f_{avg} = \frac{1}{3} \int_{- 1}^{2} x^{3} d x .$ Yeh step kyun? Width carefully compute karo: negative subtract karne se add hota hai. Yeh classic sign trap hai (parent note ki teesri mistake dekho).

Step 2. $x^{3}$ ka antiderivative $\frac{x ^{4}}{4}$ hai: $\int_{- 1}^{2} x^{3} d x = [\frac{x ^{4}}{4}]_{- 1}^{2} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} .$ Yeh step kyun? $(- 1)^{4} = + 1$ hai, $- 1$ nahi — even power sign ko khatam kar deta hai. Ise dhyan se dekho.

Step 3. $3$ se divide karo: $f_{avg} = \frac{15/4}{3} = \frac{15}{12} = \frac{5}{4} = 1.25.$ Yeh step kyun? Integral $\frac{15}{4}$ signed area hai (positive lobe minus negative lobe); width $b - a = 3$ se divide karne par woh bachi hui area average height mein convert hoti hai. Answer: $f_{avg} = \frac{5}{4} = 1.25$ .

Verify: positive hai (bada positive lobe jeet jaata hai), aur modest hai — hamare $1$ ke forecast ke karib. Yeh $min f (- 1) = - 1$ aur $max f (2) = 8$ ke beech mein hai ✓. ✓

C5 — interval sitting in negative $x$ -territory

Worked example Example 5 —

$f (x) = e^{x}$ on $[- 2, 0]$ Forecast: $e^{x}$ hamesha positive hai lekin jab $x$ negative ho toh chota hota hai: $e^{- 2} \approx 0.135$ se $e^{0} = 1$ tak. Curve low values ke paas rehti hai (yeh badhti hai, toh low pe zyada time spend karti hai), toh guess karo average values ke midpoint $(0.135 + 1) /2 \approx 0.57$ se neeche hoga, shayad around $0.43$ .

Step 1. Formula, $a = - 2, b = 0$ , width $= 0 - (- 2) = 2$ : $f_{avg} = \frac{1}{2} \int_{- 2}^{0} e^{x} d x .$ Yeh step kyun? Interval completely negative $x$ mein hai, lekin width $b - a$ phir bhi positive hai — bas yahi matter karta hai.

Step 2. $e^{x}$ ka antiderivative $e^{x}$ hai (yeh apna khud ka antiderivative hai — yahi woh poora reason hai ki $e$ special kyun hai): $\int_{- 2}^{0} e^{x} d x = [e^{x}]_{- 2}^{0} = e^{0} - e^{- 2} = 1 - e^{- 2} .$ Yeh step kyun? Humne $e^{x}$ isliye use kiya kyunki yeh woh ek function hai jo khud mein integrate hota hai, toh koi nayi cheez nahi aati.

Step 3. $2$ se divide karo: $f_{avg} = \frac{1 - e ^{- 2}}{2} = \frac{1 - 0.1353 \dots}{2} \approx \frac{0.8647}{2} \approx 0.4323.$ Yeh step kyun? Integral $1 - e^{- 2}$ curve ke neeche ki area hai; width $b - a = 2$ se divide karne par woh area evenly spread hoti hai aur average height milti hai. Answer: $f_{avg} = \frac{1 - e ^{- 2}}{2} \approx 0.432$ .

Verify: $e^{- 2} \approx 0.135$ aur $1$ ke beech mein hai ✓, aur forecast ki tarah $0.57$ se neeche hai (concave-up curve low values ko favour karti hai). ✓

C6 — degenerate case $b - a \to 0$

Worked example Example 6 —

$f (x) = x^{2}$ ka average $[2, 2 + h]$ pe jab $h \to 0$ Forecast: agar tum ek function ko $x = 2$ ke around ek tiny interval pe average karo, toh basically tum $x = 2$ pe uski value padh rahe ho. Toh guess karo limit $f (2) = 4$ hai.

Step 1. $[2, 2 + h]$ pe average likho: $f_{avg} (h) = \frac{1}{h} \int_{2}^{2 + h} x^{2} d x .$ Yeh step kyun? Yahan width $h$ hai, jise hum $0$ tak shrink hone denge. Directly $h = 0$ plug karne par $\frac{0}{0}$ milta hai — undefined — toh humhe ek limit chahiye, exactly woh tool jo "yeh kis value ke paas jaata hai" ke liye hai.

Step 2. Integral evaluate karo: $\int_{2}^{2 + h} x^{2} d x = [\frac{x ^{3}}{3}]_{2}^{2 + h} = \frac{( 2 + h ) ^{3} - 8}{3} .$ Yeh step kyun? Hum pehle integral compute karte hain ( $h$ ko symbol rakhte hue) taaki baad mein hum denominator ke troublesome $h$ ko algebraically cancel kar sakein — jise tumne likha nahi hai usse cancel nahi kar sakte.

Step 3. $(2 + h)^{3} = 8 + 12 h + 6 h^{2} + h^{3}$ expand karo, toh integral $= \frac{12 h + 6 h ^{2} + h ^{3}}{3}$ hai. $h$ se divide karo: $f_{avg} (h) = \frac{12 h + 6 h ^{2} + h ^{3}}{3 h} = \frac{12 + 6 h + h ^{2}}{3} .$ Yeh step kyun? $h$ cancel karne se $0/0$ problem dur hoti hai — ab hum safely $h = 0$ set kar sakte hain.

Step 4. $h \to 0$ lo: $lim_{h \to 0} f_{avg} (h) = \frac{12 + 0 + 0}{3} = 4.$ Yeh step kyun? Ab ki expression ek clean polynomial hai jisme $h$ se division nahi hai, $h \to 0$ lene par bas $h = 0$ substitute hota hai — yeh woh limit hai jo "window ko single point $x = 2$ tak shrink karo" capture karta hai. Answer: limiting average $4 = f (2)$ hai.

Verify: exactly forecast ki tarah. Yeh Mean Value Theorem (Derivatives) / MVT for integrals se deep link hai: jab window collapse hoti hai, average value function ki actual value us point pe collapse ho jaati hai. ✓

C7 — real-world word problem (tum $f, a, b$ build karte ho)

Worked example Example 7 — ek din mein average temperature

Ek greenhouse mein $12$ -hour day mein temperature (°C) is model se diya gaya hai $T (t) = 18 + 6 sin (\frac{π t}{12}), 0 \leq t \leq 12,$ jahan $t$ hours mein hai. Din bhar ka average temperature kya hai?

Forecast: $sin$ term temperature ko ek base of $18$ ke around upar neeche wiggle karta hai. Half-period ( $0$ to $12$ is $sin$ ke $24$ -hour period ka half hai) pe sine ek positive hump hai, toh average $18$ se upar hoga. Guess karo around $22$ .

Step 1. Pieces identify karo: $f = T$ , $a = 0$ , $b = 12$ , width $= 12$ : $T_{avg} = \frac{1}{12} \int_{0}^{12} [18 + 6 sin (\frac{π t}{12})] d t .$ Yeh step kyun? "Average over the day" word hi average-value formula hai — tumhe bas $a, b, f$ spot karna tha.

Step 2. Integral split karo (linearity): $\int_{0}^{12} 18 d t = 18 \cdot 12 = 216.$ Sine part ke liye, $sin (k t)$ ka antiderivative $- \frac{1}{k} cos (k t)$ hai, $k = \frac{π}{12}$ ke saath: $\int_{0}^{12} 6 sin (\frac{π t}{12}) d t = 6 \cdot [- \frac{12}{π} cos (\frac{π t}{12})]_{0}^{12} = - \frac{72}{π} (cos π - cos 0) .$ Yeh step kyun? Chain-rule factor $\frac{12}{π}$ chahiye kyunki andar $\frac{π t}{12}$ hai, sirf $t$ nahi.

Step 3. Evaluate karo: $cos π - cos 0 = - 1 - 1 = - 2$ , toh sine integral $= - \frac{72}{π} (- 2) = \frac{144}{π}$ hai. Yeh step kyun? Hum endpoints $t = 12$ aur $t = 0$ ko $\frac{π t}{12}$ mein plug karte hain, jo $cos π$ aur $cos 0$ deta hai; yeh exact known values $- 1$ aur $+ 1$ hain, toh bracket $- 2$ ban jaata hai aur double negative sign ko positive $\frac{144}{π}$ mein flip kar deta hai.

Step 4. Total integral $= 216 + \frac{144}{π}$ ; $12$ se divide karo: $T_{avg} = \frac{216 + \frac{144}{π}}{12} = 18 + \frac{12}{π} \approx 18 + 3.82 = 21.82 °C .$ Yeh step kyun? Integral $216 + \frac{144}{π}$ accumulated total "temperature-hours" hai; width $b - a = 12$ hours se divide karne par woh evenly spread hoti hai aur average temperature milta hai — total ÷ length, formula ka dil. Answer: $T_{avg} = 18 + \frac{12}{π} \approx 21.8$ °C.

Verify: forecast ki tarah base $18$ se upar ✓, aur actual temperature range $[18, 24]$ ke andar (kyunki $sin$ max $1$ pe hai, jo $24$ deta hai) ✓. Units: °C, sahi — $\frac{1}{12}$ ke units $1/ hr$ hain aur $d t$ ke $hr$ cancel ho jaate hain. ✓

C8 — exam twist: guaranteed point $c$ dhundho (MVT)

Worked example Example 8 — MVT for Integrals se

$c$ locate karo $f (x) = x^{2}$ on $[1, 4]$ ke liye (same as Example 1), Mean Value Theorem for Integrals (MVT) promise karta hai koi $c \in [1, 4]$ jahan $f (c) = f_{avg}$ ho. $c$ dhundho.

Forecast: Ex 1 mein humne $f_{avg} = 7$ paya. Toh humhe $c^{2} = 7$ chahiye, matlab $c = 7 \approx 2.65$ . Yeh $[1, 4]$ ke andar land karna chahiye.

Step 1. Humhare paas Example 1 se $f_{avg} = 7$ already hai. $f (c)$ ko usके barabar set karo: $c^{2} = 7.$ Yeh step kyun? MVT for integrals kehta hai kisi interior $c$ ke liye $f (c) = f_{avg}$ hoga — yeh woh equation hai jo ise dhundhti hai. Yeh isliye exist karta hai kyunki $7$ , $min = 1$ aur $max = 16$ ke beech mein hai, aur Intermediate Value Theorem guarantee karta hai ki ek continuous $f$ beech ki har value hit karti hai.

Step 2. Solve karo: $c = \pm 7$ . $c = - 7$ reject karo kyunki woh $[1, 4]$ ke bahar hai: $c = 7 \approx 2.6458.$ Yeh step kyun? Cases matter karte hain — algebra do roots deta hai, lekin sirf interval ke andar wala MVT point hai.

Answer: $c = 7 \approx 2.65$ .

Verify: $2.65 \in [1, 4]$ ✓ aur $(7)^{2} = 7 = f_{avg}$ ✓. Geometrically: yeh woh $x$ hai jahan curve flat average-line ko cross karti hai. ✓

Recall

Recall Average value negative kab hoti hai?

Jab signed area negative ho — matlab axis ke neeche ka region axis ke upar se bhaari ho. ::: $f_{avg}$ ka sign $\int_{a}^{b} f d x$ ka sign hai, kyunki $b - a > 0$ hamesha hota hai.

Recall Average exactly zero kyun ho sakta hai jab

$f$ interval pe kabhi zero nahi hoti? Positive aur negative signed areas cancel ho jaate hain. ::: Example: $sin x$ on $[0, 2 π]$ — dono humps mirror images hain, toh unka sum $0$ hota hai.

Recall Jab interval collapse hota hai (

$b - a \to 0$ ek point $p$ ke around), average kya approach karta hai? Value $f (p)$ . ::: $\frac{0}{0}$ ek limit se resolve hota hai; tiny-window average point value padhta hai.

Connections

Average value of a function — parent note; yeh page uska worked-example annex hai.
Definite Integral as Area — signed area yahan har case ka engine hai.
Riemann Sums — formula yahan se aaya tha.
Fundamental Theorem of Calculus — har integral evaluate karne ka tarika.
Intermediate Value Theorem — kyun C8 mein MVT point $c$ zaroor exist karta hai.
Mean Value Theorem (Derivatives) — sibling; C6 collapse-to-a-point link dikhata hai.
Average and Instantaneous Velocity — C7 ka word-problem flavour.