Exercises — Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs

Yeh ek constant limits wala definite integral hai → Part 2 (aur $f(x)=x^2$ continuous hai, to use karna allowed hai). Hume ek antiderivative $G$ chahiye jisme $G^{\prime }(x)=x^2$ ho. Power rule ko reverse karke ($\frac{d}{dx}{x}^n=n{x}^{n-1}$, to undo karne ke liye power mein ek add karo aur divide karo): $G(x)=\frac{x^3}{3},G^{\prime }(x)=\frac{3x^2}{3}=x^2.✓$ Ab endpoints plug in karo (top minus bottom): ${\int }_1^4{x}^{2}dx=G(4)-G(1)=\frac{64}{3}-\frac{1}{3}=\frac{63}{3}=21.$

Upper limit exactly $x$ hai (naki $x^2$, naki $3x$), aur $\sqrt{1+t^4}$ continuous hai, isliye Part 1 directly apply hota hai. Part 1 kehta hai: ek area-so-far function differentiate karne par integrand wapas milta hai, top limit par evaluate hoke. $\frac{d}{dx}{\int }_0^x\sqrt{1+t^4}dt=\sqrt{1+x^4}.$ Dhyaan do humne antiderivative nahi nikali — $\sqrt{1+t^4}$ ka koi elementary antiderivative hai hi nahi. Yahi Part 1 ka poora point hai: yeh "area accumulation ki rate kya hai?" ka jawab deta hai bina rectangles sum kiye.

Limits equal hain, to hum zero width ke ek slab ka area pooch rahe hain. ${\int }_c^c\cos tdt=0.$ Part 2 se bhi obvious hai: kisi bhi antiderivative $G$ ke liye $G(c)-G(c)=0$. Yeh chhota sa fact woh "base value" hai jo Part 1 mein $F(a)={\int }_a^{a}f=0$ banata hai.

Part 2 (aur $\cos x$ continuous hai). Hume $G$ chahiye jisme $G^{\prime }=\cos x$ ho. Kyunki $\frac{d}{dx}\sin x=\cos x$, to $G(x)=\sin x$ lo. ${\int }_0^{\pi /2}\cos xdx=\sin \frac{\pi }{2}-\sin 0=1-0=1.$

Integral ko pieces mein tod do (integrals sums mein add hote hain). $\frac{1}{x^2}=x^{-2}$ rewrite karo. Dhyaan do $3x^2-x^{-2}$ $[1,2]$ par continuous hai (beech mein koi zero nahi), isliye Part 2 legal hai.

• $3x^2$ ka antiderivative hai $x^3$.
• $-x^{-2}$ ka antiderivative: power rule use karo "power mein ek add karo, naye power se divide karo." Naya power hai $-2+1=-1$, to $\int -x^{-2}dx=-\frac{x^{-1}}{-1}=-(x^{-1}\cdot (-1))=x^{-1}=\frac{1}{x}.$ Do minus signs (ek coefficient se, ek $-1$ se divide karne se) cancel ho jaate hain. Check karo: $\frac{d}{dx}\frac{1}{x}=\frac{d}{dx}{x}^{-1}=-x^{-2}.✓$ To $G(x)=x^3+\frac{1}{x}$. ${\int }_1^2=G(2)-G(1)=(8+\frac{1}{2})-(1+1)=8.5-2=\frac{13}{2}.$

$x^3$ ka antiderivative hai $\frac{x^4}{4}$. ${\int }_{-1}^1{x}^{3}dx=\frac{1^4}{4}-\frac{(-1{)}^4}{4}=\frac{1}{4}-\frac{1}{4}=0.$ Zero kyun? Figure dekho. $x^3$ ek odd function hai: $-1$ se $0$ tak ka piece axis ke neeche hai (negative signed area) aur $0$ se $1$ tak ke upar wale piece ka mirror image hai (positive). Woh exactly cancel karte hain.

Top limit $x^2$ hai, naki $x$, isliye Part 1 akela kaafi nahi — yeh bhi account karna hoga ki limit khud kitni tezi se move kar rahi hai. $u=x^2$ aur $H(u)={\int }_0^u\sin tdt$ set karo. Part 1 se ($\sin t$ continuous), $H^{\prime }(u)=\sin u$. Chain rule se (inner variable ke through rate): $\frac{d}{dx}H(x^2)=H^{\prime }(x^2)\cdot \frac{d}{dx}(x^2)=\sin (x^2)\cdot 2x.$ Yahan earn kiya general rule: $\frac{d}{dx}{\int }_a^{g(x)}f(t)dt=f(g(x))g^{\prime }(x)$.

Variable lower limit mein hai. Pehle limits flip karo (limits swap karne se sign flip hota hai): ${\int }_x^5{e}^{-t^2}dt=-{\int }_5^x{e}^{-t^2}dt.$ Ab top limit $x$ hai, aur $e^{-t^2}$ continuous hai, isliye Part 1 deta hai: $\frac{d}{dx}\left(-{\int }_5^x{e}^{-t^2}dt\right)=-e^{-x^2}.$ Takeaway: moving lower limit minus sign ke saath contribute karta hai.

Dono limits move kar rahi hain. Kisi constant $k$ par split karo (interval additivity), jahan $k$ bas koi fixed number hai jo hum "meeting point" ke roop mein choose karte hain: ${\int }_{x^2}^{x^3}={\int }_k^{x^3}-{\int }_k^{x^2}.$ Har piece par L3.1 wala general rule apply karo ($f(t)=\ln (1+t^2)$, jo continuous hai): $\frac{d}{dx}=\ln (1+(x^3{)}^2)\cdot 3x^2-\ln (1+(x^2{)}^2)\cdot 2x$ $=3x^2\ln (1+x^6)-2x\ln (1+x^4).$ Answer $k$ par kyun depend nahi karta: split karne se constant pieces ${\int }_k^{x^3}$ aur ${\int }_k^{x^2}$ introduce hote hain, lekin har $k$-flavoured contribution lower limit $k$ mein bake ho jaati hai, jo ek constant hai — aur constant limit ka derivative $0$ hota hai. To $k$ change karne se dono integrals ek fixed number se shift ho jaate hain jo differentiate ho ke khatam ho jaata hai. Sirf moving limits $x^3$ aur $x^2$ survive karte hain. Top limit add karta hai; bottom limit subtract karta hai — har ek apna chain-rule factor lekar.

Part 1 se (integrand $t^2-4t+3$ continuous hai), $F^{\prime }(x)=x^2-4x+3=(x-1)(x-3)$. Critical points jahan $F^{\prime }=0$: $x=1$ aur $x=3$. $F^{\prime }(x)=(x-1)(x-3)$ ka sign chart:

• $x<1$: dono factors negative → $F^{\prime }>0$ (rising).
• $1<x<3$: $(x-1)>0,(x-3)<0$ → $F^{\prime }<0$ (falling).
• $x>3$: dono positive → $F^{\prime }>0$ (rising).

To $F$ pehle rise karta hai, phir fall, phir rise: $x=1$ par local maximum, $x=3$ par local minimum.

Jab $x\to 0$ to integral $\to 0$ aur $x\to 0$, yeh indeterminate form $\frac{0}{0}$ deta hai. Do clean routes hain:

Route A — derivative pehchano. Maano $A(x)={\int }_0^x\cos (t^2)dt$, to $A(0)=0$ aur Part 1 se ($\cos (t^2)$ continuous) $A^{\prime }(x)=\cos (x^2)$. Tab ${\lim }_{x\to 0}\frac{A(x)}{x}={\lim }_{x\to 0}\frac{A(x)-A(0)}{x-0}=A^{\prime }(0)=\cos (0)=1.$ Yoh limit hai hi $A$ ka $0$ par derivative ki definition.

Route B — L'Hôpital. $\frac{0}{0}$ form, top aur bottom differentiate karo: top $\to \cos (x^2)$ (Part 1), bottom $\to 1$. Limit $=\cos (0)=1$. $\boxed{\underset{x\to 0}{\lim }\frac{1}{x}{\int }_0^x\cos (t^2)dt=1.}$

Integrand mein $x=1$ par ek corner hai jahan $||$ ke andar ka sign change karta hai. Corner par split karo aur absolute value ko har piece par sahi sign ke saath drop karo:

• $[0,1]$ par: $x-1\le 0$, to $|x-1|=1-x$.
• $[1,2]$ par: $x-1\ge 0$, to $|x-1|=x-1$.

${\int }_0^1(1-x)dx=[x-\frac{x^2}{2}{]}_0^1=(1-\frac{1}{2})-0=\frac{1}{2}.$ ${\int }_1^2(x-1)dx=[\frac{x^2}{2}-x{]}_1^2=(2-2)-(\frac{1}{2}-1)=0-(-\frac{1}{2})=\frac{1}{2}.$ Total $=\frac{1}{2}+\frac{1}{2}=1$. (Geometrically: do right triangles, har ek ka area $\frac{1}{2}$.)

Dono sides differentiate karo. Left side ek area-so-far function hai top limit $x$ ke saath (aur $f$ continuous di hui hai), to Part 1 deta hai $\frac{d}{dx}{\int }_0^{x}f=f(x)$. Right side ko product rule chahiye: $\frac{d}{dx}\left(x{e}^{2x}\right)=e^{2x}+x\cdot 2e^{2x}=e^{2x}(1+2x).$ Isliye $f(x)=e^{2x}(1+2x).$ $x=0$ par: $f(0)=e^0(1+0)=1.$

Ek ek layer utaro. Maano $g(u)={\int }_0^u\frac{1}{1+s^2}ds$; integrand $\frac{1}{1+s^2}$ continuous hai. Part 1 se, $F^{\prime }(x)=g(x)={\int }_0^x\frac{1}{1+s^2}ds.$ Dobara differentiate karo, $g$ par Part 1 use karke: $F^{\prime \prime }(x)=g^{\prime }(x)=\frac{1}{1+x^2}.$ Isliye $F^{\prime \prime }(0)=\frac{1}{1+0}=1.$ (Do nested integrals ⇒ Part 1 ke do applications unhe strip karne ke liye.)

$f(x)=x$ lo (ya koi bhi odd continuous function). Tab ${\int }_{-1}^{1}xdx=[\frac{x^2}{2}{]}_{-1}^1=\frac{1}{2}-\frac{1}{2}=0,$ phir bhi $f(x)=x$ zero function nahi hai. Part 1 se contradiction kyun nahi: Part 1 variable-limit function $F(x)={\int }_{-1}^{x}f$ ke derivative ki baat karta hai, kehta hai $F^{\prime }(x)=f(x)$. Yeh kuch bhi force nahi karta ki ek fixed interval par koi single definite integral nonzero ho. Signed area cancel hokar $0$ de sakta hai jabki height $f$ genuinely nonzero ho.

$f$ continuous) Constant-limit definite integral ::: ${\int }_a^{b}f=G(b)-G(a)$ (Part 2), koi bhi antiderivative $G$. Top limit $x$ hai ::: $\frac{d}{dx}{\int }_a^{x}f=f(x)$ (Part 1). Top limit $g(x)$ hai ::: $\frac{d}{dx}{\int }_a^{g(x)}f=f(g(x))g^{\prime }(x)$ (Part 1 + Chain Rule). Dono limits move karti hain ::: $\frac{d}{dx}{\int }_{p(x)}^{q(x)}f=f(q)q^{\prime }-f(p)p^{\prime }$ (ek constant par split karo).