04.05.08.නිශ්චිත අනුකලනය

අර්ථ දැක්වීම

\frac{\operatorname d\varnothing(x)}{\operatorname dx}=f(x)\;\text{නම්}\;\int f(x)\;dx\;=\varnothing(x)\;+\;C\;\text{වේ.}\;\text{එය අනිශ්චිත අනුකලනය ලෙස හැදින්වේ. }\;

\int_a^bf\left(x\right)\operatorname dx=\left[\phi\left(x\right)\right]_a^b=\phi\left(b\right)-\phi\left(a\right)
ලෙස ලියනු ලැබේ.

• මෙවැනි අනුකලනයක් නිශ්චිත අනුකලනයක් ලෙස හැඳින්වේ.
• මෙහි a යටත් සීමාව වන අතර b උඩත් සීමාව වේ.

උදා:

1.\begin{array}{rcl}\int_1^2x^2dx\;&=&\;\left[\frac{x^3}3\right]_1^2\\&=&\;\frac13(2^3\;-\;1^3)\\&=&\frac73\end{array}

2.\int_1^2\frac x{\sqrt{x^2+1}}dx
\frac d{dx}\left(\sqrt{x^2+1}\right)=\frac12.\frac1{\sqrt{x^2+1}}.2x=\;\frac x{\sqrt{x^2+1}}\Rightarrow\int\frac{\displaystyle x}{\displaystyle\sqrt{x^2+1}}dx=\sqrt{x^2+1}
\begin{array}{rcl}\int_0^\sqrt3\frac x{\sqrt{x^2+1}}\operatorname dx &=&\left[\sqrt{x^2+1}\right]_0^\sqrt3\\\;&=&\sqrt{\left(\sqrt3\right)^2+1}-\sqrt{0+1}\\\;&=&2-1\\\;&=&1\end{array}

3. \begin{array}{rcl}\int_1^4\frac{e^\sqrt x}{\sqrt x}\operatorname dx &=&2\int_1^4\frac{e^\sqrt x}{2\sqrt x}\operatorname dx \\\;&=&2\left[e^\sqrt x\right]_1^4\\\;&=&2\left(e^\sqrt4-e^\sqrt1\right)\\\;&=&2\left(e^2-e\right)\end{array}

4.\int_0^ 1\sqrt{4-x^2}\;dx
මෙය විසඳීමට පෙර පරිදී   x = 2sinθ යොදා ගනී.
අනිශ්චිත අනුකලනයේදී අවකලනය කර පසුව ආදේශ කෙරේ. නමුත් මෙහිදී ඊට අමතරව විචල්‍ය මාරු කෙරෙන නිසා උඩත් හා යටත් සීමා වෙනස් වේ. එම නිසා ඒවාද සොයා ගත යුතුයි.
\begin{array}{l}x\;=\;2\sin\theta\\x\;=\;1\;\text{විට }\;\frac12\;=\;\sin\theta\;\Rightarrow\theta\;=\;\frac{\mathrm\pi}6\\x\;=\;0\;\mathrm{විට}\;0\;=\;\sin\theta\;\;\;\;\Rightarrow\;\theta\;=\;0\\\frac{\operatorname dx}{\operatorname d\theta}\;=\sin\theta\;\;\Rightarrow\;dx\;=\;\sin\theta d\theta\end{array}
\begin{array}{l}\begin{array}{rcl}\int_0^1\sqrt{4-x^2}\;dx\;&=&\;\int_0^\frac\pi6(\sqrt{4-4\sin^2\theta})\sin\theta\;d\theta\\&=&\int_0^\frac\pi6(\sqrt{4-4\sin^2\theta})\sin\theta\;d\theta\\&=&\int_0^\frac\pi62\sqrt{\cos^2\theta}\sin\theta\;d\theta\\&=&\int_0^\frac\pi62\cos\theta\sin\theta\;d\theta\\&=&\int_0^\frac\pi6\sin2\theta\;d\theta\\&=&-\frac12\left[\cos2\theta\right]_0^\frac\pi6\\&=-&\frac12\cos2\left(\frac{\mathrm\pi}6\right)\;-\;\cos0\\&=&-\frac12(\frac12-1)\\&=&\frac14\end{array}\\\\\end{array}

5.\int_0^\frac\pi2\frac{dx}{3\cos x-4\sin x+5}
\begin{array}{rcl}tan\frac x2&=&t\;\text{ආදේශය }x=2tan^{-1}t\\\frac{dx}{dt}&=&\frac2{1+t^2}\Longrightarrow dx=\frac{2dt}{1+t^2}\\x&=&0\text{විට}t=tan0=0\\x&=&\frac\pi2\text{විට }t=tan\frac\pi4=1\\cosx&=&\frac{1-\tan^2{\displaystyle\frac x2}}{1+tan^2\frac x2}=\frac{1-t^2}{1+t^2}\\sinx&=&\frac{2\tan{\displaystyle\frac x2}}{1+tan^2\frac x2}=\frac{2t}{1+t^2}\\&&\end{array}
\begin{array}{rcl}&&\\\int_0^\frac\pi2\frac{dx}{3\cos x-4\sin x+5}&=&\int_0^1\frac{\displaystyle\frac{2dt}{1+t^2}}{3(\frac{1-t^2}{1+t^2})-4(\frac{2t}{1+t^2})+5}\\&=&\int_0^1\frac{\displaystyle\frac{2dt}{1+t^2}}{3(\frac{1-t^2}{1+t^2})-4(\frac{2t}{1+t^2})+5}\\&=&\int_0^1\frac{2dt}{3(1-t^2)-4(2t)+5(1+t^2)}\\&=&\int_0^1\frac{dt}{t^2-4t+4}\\&=&\int_0^1\frac{dt}{(t-2)^2}\\&=&\lbrack-(t-2)^{-1}\rbrack_0^1\\&=&-\frac1{(-1)}-(\frac{-1}{-2})\\&=&\frac12\\&&\end{array}

නිශ්චිත අනුකලනයේ ලක්ෂණ

\begin{array}{c}1.\int_a^bf(x)dx=\int_a^bf(t)dt\\x=t\\dx=dt\\x\rightarrow a\;\text{විට }t\rightarrow a\\x\rightarrow b\;\text{විට }t\rightarrow b\\\int_a^bf(x)dx=\int_a^bf(t)dt\end{array}
\begin{array}{c}2.\int_a^bf(x)dx=-\int_b^af(x)dx\\{\text{සාධනය}:-f(x)\text{හ ිඅනුකල්‍ය }F(x)\text{යැයි ගනිමු}.}\\\int_a^bf(x)dx=\lbrack F(x)\rbrack_a^b\;\;\\=F(b)-F(a)\\=-\lbrack F(a)-F(b)\rbrack\\=-\lbrack F(x)\rbrack_b^a\;\;\;\;\\=-\int_b^af(x)dx\end{array}
\begin{array}{rcl}3.\int_a^bf(x)dx&=&\int_a^cf(x)dx+\int_c^bf(x)dx\\\text{සාධනය}:\int_a^bf(x)dx&=&\lbrack F(x){\rbrack_a^b\;}\\&=&F(b)-F(a)\\&=&F(b)-F(c)+F(c)-F(a)\\&=&\lbrack F(x)\rbrack_c^b+\lbrack F(x)\rbrack_a^c\\&=&\int_c^bf(x)dx+\int_a^cf(x)dx\\&=&\int_a^cf(x)dx+\int_c^bf(x)dx\end{array}
\begin{array}{rcl}4.\int_0^af(x)dx&=&\int_0^af(a-x)dx\\\text{ආදේශය }\\x&=&a-t\;\\dx&=&-dt\;\\x&\rightarrow&0\;\text{විට }t\rightarrow a\\x&\rightarrow&a\;\text{විට }t\rightarrow0\\\int_0^af(x)dx&=&-\int_a^0f(a-t)dt\\&=&\int_0^af(a-t)dt=\int_0^af(a-x)dx;\text{පළමු ලක්ෂණයට අනුව }\end{array}
\begin{array}{rcl}5.\int_a^bf(x)dx&=&\int_a^bf(a+b-x)dx\\\text{ආදේශය }\\x&=&a+b-t\\dx&=&-dt\\x&\rightarrow&a\;\text{විට }t\rightarrow b\\x&\rightarrow&b\;\text{විට }t\rightarrow a\\\int_a^bf(x)dx&=&-\int_b^af(a+b-t)dt\\&=&\int_a^bf(a+b-t)dt;\;\;\;\;\text{පළමු ලක්ෂණයට අනුව}\\&=&\int_a^bf(a+b-x)dx\end{array}

උදා:-

1.(04) ලක්ෂණය භාවිතයෙන් I හි අගය සොයන්න
\begin{array}{rcl}I&=&\int_0^{\pi/2}\frac{Sin^3x}{Cos^3x+Six^3x}dx------(1)\\I&=&\int_0^{\pi/2}\frac{Sin^3({\displaystyle\frac\pi2}-x)dx}{Cos^3(\frac\pi2-x)+Sin^3(\frac\pi2-x)}\\I&=&\int_0^{\pi/2}\frac{Cos^3x}{Sin^3x+Cos^3x}dx------(2)\\&&(1)+(2)\\2I&=&\int_0^{\pi/2}\frac{Sin^3x+Cos^3x}{Sin^3x+Cos^3x}dx\\&=&\int_0^{\pi/2}dx\\&&\\&=&\lbrack x\rbrack_0^{\pi/2}\\&&\\&=&\frac\pi2-0\\&=&\frac\pi2\\I&=&\frac\pi4\\&&\end{array}
2.\begin{array}{rcl}&&\int_0^\pi\frac{x\sin xdx}{1+cos^2x}dx\\I&=&\int_0^\pi\frac{x\sin xdx}{1+cos^2x}dx-----(1)\\I&=&\int_0^\pi\frac{(\mathrm\pi-x)\sin{(\mathrm\pi-x)}dx}{1+cos^2{(\mathrm\pi-x)}}\\I&=&\int_0^\pi\frac{(\mathrm\pi-x)\sin{(x)}dx}{1+cos^2{(x)}}\;-----(2)\\&&(1)+(2)\\2I&=&\int_0^\pi\frac{x\sin xdx}{1+cos^2x}dx\;+\;\int_0^\pi\frac{(\mathrm\pi-x)\sin{(x)}dx}{1+cos^2{(x)}}\\2I\;&=&\;\int_0^\pi\frac{\mathrm{πsin}xdx}{1+cos^2x}dx\\I&=&\frac{\mathrm\pi}2\int_0^\pi\frac{\sin xdx}{1+cos^2x}dx\\\frac{d\cos x}{dx}\;&=&\;-\sin x\;\Rightarrow d\cos x\;=\;-\sin xdx\\\;I&=&\frac{\mathrm\pi}2\int_0^\pi\frac{-d\cos x}{1+cos^2x}dx\\I&=&-\frac{\mathrm\pi}2\int_0^\pi\frac{d\cos x}{1+(cosx)^2}dx\\&=&-\frac{\mathrm\pi}2\lbrack tan^{-1}(cos\pi)\rbrack_0^{\mathrm\pi}\\I&=&-\frac{\mathrm\pi}2\lbrack tan^{-1}(cos\pi)-tan^{-1}(\cos0)\rbrack\\I&=&{-\frac{\mathrm\pi}2\lbrack tan^{-1}(-1)-tan^{-1}(1)\rbrack\;}\\I&=&-\frac{\mathrm\pi}2\lbrack\;\frac{\mathrm\pi}4-(-\frac{\mathrm\pi}4)\rbrack\\I&=&\frac{\mathrm\pi^2}4\end{array}

 “The theory of probabilities is at the bottom nothing but common sense reduced to calculus”
-Pierre Laplace –