Exercises — Average value of a function

Ek constant function height $7$ par ek flat line hai. Kisi bhi base ke upar us function ke neeche ka region height $7$ ka rectangle hai, isliye clearly average height $7$ hai. Formally: $f_{\text{avg}}=\frac{1}{10-2}{\int }_2^{10}7dx=\frac{1}{8}[7x{]}_2^{10}=\frac{1}{8}\cdot 7\cdot (10-2)=\frac{56}{8}=7.$ $7x$ kyun? Constant $7$ ka antiderivative $7x$ hai, kyunki $\frac{d}{dx}(7x)=7$; $7x$ ko $2$ se $10$ tak evaluate karne par $7\cdot (10-2)$ milta hai. Answer: $7$. Ek constant ka average woh khud hota hai — picture already ek rectangle hai.

(b). Integral ${\int }_a^{b}fdx$ ke units hain (height of $f$) $\times$ (length along $x$) — yeh ek area hai, height nahi. Average height paane ke liye width $b-a$ se divide karke length hatani padegi. Answer: (b), kyunki average = area ÷ width.

Figure dekho: straight line evenly $(0,0)$ se $(10,10)$ tak chadhti hai. Neeche wala blue triangle beech se slice karke flip kiya ja sakta hai — triangle ka upar wala aadha hissa height $5$ par dashed line wale rectangle ke khali top-left corner ko exactly bhar deta hai. Isliye line height $5$ ke neeche aur upar "barabar time bitaati" hai, aur average midpoint value par baithta hai. $[0,10]$ ka midpoint $5$ hai, aur $f(5)=5$. $f_{\text{avg}}=\frac{1}{10}{\int }_0^{10}xdx=\frac{1}{10}{\left[\frac{x^2}{2}\right]}_0^{10}=\frac{1}{10}\cdot \frac{10^2}{2}=\frac{1}{10}\cdot 50=5.$ $x^2/2$ kyun? Power rule: $\int xdx=x^2/2$, kyunki $\frac{d}{dx}\left(\frac{x^2}{2}\right)=x$. Answer: $5$. Kisi bhi straight line ke liye average midpoint par $f$ ki value ke barabar hota hai.

Width hai $b-a=4-1=3$. $f_{\text{avg}}=\frac{1}{3}{\int }_1^4{x}^{2}dx=\frac{1}{3}{\left[\frac{x^3}{3}\right]}_1^4.$ $x^3/3$ kyun? Power rule: $\int x^{2}dx=x^3/3$ (dekho Fundamental Theorem of Calculus). $=\frac{1}{3}\cdot \frac{4^3-1^3}{3}=\frac{1}{3}\cdot \frac{64-1}{3}=\frac{1}{3}\cdot \frac{63}{3}=\frac{63}{9}=7.$ Answer: $7$.

Width hai $\pi /2-0=\pi /2$. $f_{\text{avg}}=\frac{1}{\pi /2}{\int }_0^{\pi /2}\cos xdx=\frac{2}{\pi }[\sin x{]}_0^{\pi /2}.$ $\sin x$ kyun? Kyunki $\frac{d}{dx}\sin x=\cos x$, isliye $\int \cos xdx=\sin x$. $=\frac{2}{\pi }(\sin \frac{\pi }{2}-\sin 0)=\frac{2}{\pi }(1-0)=\frac{2}{\pi }\approx \mathrm{0.637.}$ Answer: $\frac{2}{\pi }\approx 0.637$. ($[0,\pi ]$ par $\sin$ ke barabar — symmetry ka ek sundar coincidence.)

Width hai $1$. $f_{\text{avg}}=\frac{1}{1}{\int }_0^1{e}^{x}dx=[e^x{]}_0^1=e^1-e^0=e-1\approx \mathrm{1.718.}$ $e^x$ khud ko integrate kyun karta hai? $\frac{d}{dx}{e}^x=e^x$, isliye antiderivative phir se $e^x$ hai. Answer: $e-1\approx 1.718$. Dhyaan do ki yeh $f(0)=1$ aur $f(1)=e\approx 2.718$ ke beech hai, jaisa ki kisi bhi average ko hona chahiye.

Prediction: ek poore period mein, axis ke upar ka har hump neeche ke ek barabar trough se mirror hota hai. Positive area negative area ko cancel karta hai, isliye signed area $0$ hai, aur average bhi $0$. $f_{\text{avg}}=\frac{1}{2\pi }{\int }_0^{2\pi }\sin xdx=\frac{1}{2\pi }[-\cos x{]}_0^{2\pi }=\frac{1}{2\pi }(-\cos 2\pi +\cos 0)=\frac{1}{2\pi }(-1+1)=0.$ $-\cos x$ kyun? Humein ek aisi function chahiye jiska derivative $\sin x$ ho. Kyunki $\frac{d}{dx}\cos x=-\sin x$, sign flip karne par $\frac{d}{dx}(-\cos x)=\sin x$ milta hai, isliye $\int \sin xdx=-\cos x$. Answer: $0$. Average value signed area use karta hai — yahi wajah hai ki yeh zero ho sakta hai chahe function aksar nonzero ho.

Width $=8-2=6$, isliye $f_{\text{avg}}=\frac{30}{6}=5.$ Mean Value Theorem for Integrals: agar $f$, $[a,b]$ par continuous hai, toh kam se kam ek $c\in [a,b]$ exist karta hai jiske liye $f(c)=\frac{1}{b-a}{\int }_a^{b}fdx$. (Yeh Mean Value Theorem (Derivatives) ka integral-form sibling hai.) Yahan apply karte hue: kyunki $f$, $[2,8]$ par continuous hai, kam se kam ek $c\in [2,8]$ hai jiske liye $f(c)=5$. Yeh exist kyun karta hai? Average $5$, $f$ ke min aur max ke beech hai; ek continuous function Intermediate Value Theorem ke hisaab se beech ki har value tak pahunchta hai. Answer: $f_{\text{avg}}=5$; koi $c\in [2,8]$ hai jiske liye $f(c)=5$.

Flat line height $f_{\text{avg}}=3$ par hai. Iska rectangle (base $3$, height $3$) ka area $9$ hai, jo ki bilkul ${\int }_0^3{x}^{2}dx=9$ ke barabar hai.

• Left side par (pink), curve line ke neeche jaata hai: rectangle ka wahan curve se zyaada area hai — ek deficit.
• Right side par (yellow), curve line ke upar nikalti hai: curve ka rectangle se zyaada area hai — ek surplus. Kyunki rectangle aur curve dono same total area $9$ enclose karte hain, pink deficit aur yellow surplus barabar hone chahiye — ek doosre ko bhar deta hai. Yahi "average height" ka poora matlab hai. Answer: barabar areas, kyunki dono figures same total area $9$ store karti hain.

Pehle $f_{\text{avg}}=\frac{1}{3}{\int }_0^3{x}^{2}dx=\frac{1}{3}\cdot \frac{27}{3}=3$. Theorem kehta hai koi $c$ hai jiske liye $f(c)=c^2=3$, isliye $c=\pm \sqrt{3}$. Humein $c\in [0,3]$ chahiye, isliye positive root lo: $c=\sqrt{3}\approx 1.732\in [0,3].✓$ Answer: $c=\sqrt{3}\approx 1.732$. (Negative root $-\sqrt{3}$ reject hai — yeh $[0,3]$ ke bahar hai. Dono signs cover karna aur invalid ko discard karna yahi point hai.)

(i) Formula: $v_{\text{avg}}=\frac{1}{5-0}{\int }_0^{5}6tdt=\frac{1}{5}[3t^2{]}_0^5=\frac{1}{5}\cdot 3\cdot 25=\frac{75}{5}=15\text{m/s}.$ (ii) Displacement ÷ time: displacement $={\int }_0^{5}6tdt=3\cdot 25=75$ m; time $=5$ s; isliye $75/5=15$ m/s. Dono match karte hain kyunki formula disguise mein "total displacement over total time" hi hai (dekho Average and Instantaneous Velocity). Answer: dono tareekon se $15$ m/s.

$f_{\text{avg}}=\frac{1}{4}{\int }_0^{4}kxdx=\frac{1}{4}\cdot k\cdot \frac{4^2}{2}=\frac{1}{4}\cdot k\cdot 8=2k.$ $2k=6$ set karo $\Rightarrow k=3$. Answer: $k=3$. Sanity check: origin se guzarne waali line ke liye average midpoint value $f(2)=2k=6$ hoti hai. ✓

Average compute karo: $f_{\text{avg}}=\frac{1}{q-p}{\int }_p^q(mx+b)dx=\frac{1}{q-p}{\left[\frac{m{x}^2}{2}+bx\right]}_p^q.$ Bracket evaluate karo: $=\frac{1}{q-p}\left(\frac{m(q^2-p^2)}{2}+b(q-p)\right).$ $q^2-p^2=(q-p)(q+p)$ factor karo aur common $(q-p)$ cancel karo: $=\frac{m(q+p)}{2}+b=m\cdot \frac{p+q}{2}+b=f\left(\frac{p+q}{2}\right).$ Ho gaya. Straight line ke liye average value exactly midpoint height hoti hai — yahi wajah hai ki lines ke liye "endpoints average karo" ittefaqan kaam karta hai (lekin sirf lines ke liye).

$f_{\text{avg}}=\frac{1}{L}{\int }_0^L{x}^{2}dx=\frac{1}{L}\cdot \frac{L^3}{3}=\frac{L^2}{3}.$ Midpoint value hai $f(L/2)=(L/2{)}^2=L^2/4$, aur endpoints average karne par $\frac{0+L^2}{2}=\frac{L^2}{2}$ milta hai — dono mein se koi bhi $\frac{L^2}{3}$ se match nahi karta. Kyun? $x^2$ curved (convex) hai: pehle dheere badhti hai phir tezi se. Yeh apna zyaada safar chhoti values par bitaati hai, jo average ko endpoints ke midpoint se neeche kheenchta hai. Sirf straight lines mein endpoint-averaging kaam karta hai. Answer: $\frac{L^2}{3}$; curvature ise kisi bhi midpoint shortcut se alag banata hai.

$F(x)$ ko $f$ ka antiderivative maano, toh ${\int }_a^{b}f=F(b)-F(a)$. Phir $f_{\text{avg}}=\frac{F(b)-F(a)}{b-a}.$ Yeh exactly ek difference quotient hai — $a$ aur $b$ ke beech $F$ ki slope. Jab $b\to a$ toh yeh derivative ban jaata hai: ${\lim }_{b\to a}{f}_{\text{avg}}={\lim }_{b\to a}\frac{F(b)-F(a)}{b-a}=F^{\prime }(a)=f(a).$ Answer: $f_{\text{avg}}\to f(a)$. Jab tum $a$ ke aas-paas chhote se chhote window par average karte ho, toh average single value $f(a)$ par collapse ho jaata hai — bilkul samajh mein aata hai, aur yeh kaam FTC ($F^{\prime }=f$) kar raha hai. Degenerate case $b=a$ (width $0$) definition se excluded hai kyunki $\frac{0}{0}$ undefined hai; lekin limit hamein batata hai ki wahan kya value fill karein.

$f(x)=x^2$ ka average $[1,4]$ par? ::: $7$ $\cos x$ ka average $[0,\pi /2]$ par? ::: $\frac{2}{\pi }\approx 0.637$ $e^x$ ka average $[0,1]$ par? ::: $e-1\approx 1.718$ $\sin x$ ka average ek full period $[0,2\pi ]$ par? ::: $0$ (signed area cancel ho jaata hai) $x^2$ ke liye $[0,3]$ par MVT ka $c$? ::: $\sqrt{3}\approx 1.732$ $x^2$ ka average $[0,L]$ par? ::: $\frac{L^2}{3}$ $f_{\text{avg}}$ jab $b\to a$? ::: $f(a)$