$ \displaystyle y=x^3$ ကိုပေးထားတဲ့ အခါ $ \displaystyle \frac{dy}{dx}=3x^2$ လို့ရခဲ့တာ သိပြီး ဖြစ်မှာပါ။
Differentiation is the process of obtaining the derivative $ \displaystyle \frac{dy}{dx}$ from $ \displaystyle y$.
ပေးထားသော function ($ \displaystyle y$) ကနေ $ \displaystyle \frac{dy}{dx}$ ရအောင်လုပ်တဲ့ လုပ်ငန်းစဉ်ကို differentiation လုပ်တယ်လို့ ခေါ်ပါတယ်။
The reverse process of obtaining $ \displaystyle y$ from $ \displaystyle \frac{dy}{dx}$ is called integration. Therefore integration is the reverse of differentiation.
ဒါကြောင့် ...
အပြန်အလှန်အားဖြင့် $ \displaystyle \frac{dy}{dx}$ ကနေ မူလ function function ($ \displaystyle y$) ပြန်ရအောင် လုပ်တဲ့ လုပ်ငန်းစဉ်ကိုတော့ anti-derivative (integration) လို့ ခေါ်ပါတယ်။
အောက်ပါ ဥပမာများကို ဆက်ကြည့်ရအောင် ...
| $ \displaystyle y={{x}^{3}}\Rightarrow \frac{{dy}}{{dx}}=3{{x}^{2}}$ $ \displaystyle y={{x}^{3}}+2\Rightarrow \frac{{dy}}{{dx}}=3{{x}^{2}}$ $ \displaystyle y={{x}^{3}}-\frac{1}{2}\Rightarrow \frac{{dy}}{{dx}}=3{{x}^{2}}$ $\displaystyle y={{x}^{3}}+c\Rightarrow \frac{{dy}}{{dx}}=3{{x}^{2}},c =\text{constant}$ |
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ဒါကြောင့် ...
| $ \displaystyle \frac{{dy}}{{dx}}=3{{x}^{2}}\Rightarrow y=\int{{3{{x}^{2}}dx=}}{{x}^{3}}+c$ |
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ဒီနေရာမှာ $ \displaystyle c$ ကို arbitrary constant လို့ ခေါ်ပါတယ်။ $ \displaystyle \int{{3{{x}^{2}}dx}}$ ကိုတော့ indefinite integral of $ \displaystyle 3x^2$ with respect to $ \displaystyle x$ လို့ ခေါ်ပါတယ်။ indefinite လို့သုံးရတာကတော့ $ \displaystyle \int{{3{{x}^{2}}dx}}$ အတွက် အဖြေများစွာ (infinitely many solutions) ရှိနေလို့ပါပဲ။
| Definition : If $ \displaystyle {F}'(x)=f\left( x \right)$ is continuous at a given interval, then $ \displaystyle \int{{f(x)}}dx=F(x)+c$ |
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Basic Integration Rules
Integration ဥပဒေသများကို နားလည်ရန် Differentiation ဥပဒေသများနှင့် တွဲမှတ်သင့်ပါသည်။ အဘယ်ကြောင့် ဆိုသော် integration ဆိုသည်မှာ reverse process of differentiation ဖြစ်သောကြောင့်ပင်။ အောက်ပါတို့သည် integration ဆိုင်ရာ အခြေခံဥပဒေသများ ဖြစ်ပါသည်။
| Differentiation Formula | Integration Formula |
| $ \displaystyle \frac{d}{{dx}}\left[ C \right]=0,C=\text{constant}$ | $ \displaystyle \int{{0\ dx=C}},C=\text{constant}$ |
| $ \displaystyle \frac{d}{{dx}}\left[ {kx} \right]=k$ | $ \displaystyle \int{{k\ dx=kx+C}}$ |
| $ \displaystyle \frac{d}{{dx}}\left[ {kf\left( x \right)} \right]=k\ {f}'\left( x \right)$ | $\displaystyle \int{{kf\left( x \right)\ dx}}=\ \ k\int{{f\left( x \right)\ dx}}$ |
| $ \displaystyle \frac{d}{{dx}}\left[ {f\left( x \right)\pm g\left( x \right)} \right]=\frac{d}{{dx}}f\left( x \right)\pm \frac{d}{{dx}}g\left( x \right)$ | $ \displaystyle \int{{\left[ {f\left( x \right)\pm g\left( x \right)} \right]\ dx}}= \int{{f\left( x \right)\ dx}}\pm \int{{g\left( x \right)\ dx}}$ |
| $ \displaystyle \frac{d}{{dx}}\left[ {{{x}^{n}}} \right]=n{{x}^{{n-1}}}$ | $ \displaystyle \int{{{{x}^{n}}\ dx=\frac{{{{x}^{{n+1}}}}}{{n+1}}+C}},n\ne -1 $ |
| $ \displaystyle \frac{d}{{dx}}{{\left( {ax+b} \right)}^{n}}=na\left( {ax+b} \right)$ | $ \displaystyle \int{{{{{\left( {ax+b} \right)}}^{n}}}}\ dx=\frac{{{{{\left( {ax+b} \right)}}^{{n+1}}}}}{{a\left( {n+1} \right)}}+C,n\ne -1,\ a\ne 0$ |
| $ \displaystyle \frac{d}{{dx}}\left[ {\sin x} \right]=\cos x $ | $ \displaystyle \int{{\cos x\ \ dx=\sin x+C}}$ |
| $ \displaystyle \frac{d}{{dx}}\left[ {\cos x} \right]=-\sin x$ | $ \displaystyle \int{{\sin x\ \ dx=-\cos x+C}}$ |
| $ \displaystyle \frac{d}{{dx}}\left[ {\tan x} \right]={{\sec }^{2}}x$ | $ \displaystyle \int{{{{{\sec }}^{2}}x\ \ dx=\tan x+C}}$ |
| $ \displaystyle \frac{d}{{dx}}\left[ {\cot x} \right]=-{{\operatorname{cosec}}^{2}}x$ | $ \displaystyle \int{{{{{\operatorname{cosec}}}^{2}}x\ \ dx=-\cot x+C}}$ |
| $ \displaystyle \frac{d}{{dx}}\left[ {\sec x} \right]=\sec x\tan x$ | $ \displaystyle \int{{\sec x\tan x\ \ dx=\sec x+C}}$ |
| $ \displaystyle \frac{d}{{dx}}\left[ {\operatorname{cosec}x} \right]=-\operatorname{cosec}x\cot x$ | $ \displaystyle \int{{\operatorname{cosec}x\cot x\ \ dx=-\operatorname{cosec}x+C}}$ |
1.Integration of Power Functions
$\displaystyle \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\displaystyle \int{{{{x}^{n}}\ dx=\frac{{{{x}^{{n+1}}}}}{{n+1}}+C}},n\ne -1 } \\ \hline \end{array}\end{array}$
Example (1) Find the integral of each of the following.
(a) $\displaystyle x^2 $
(b) $ \displaystyle \frac{{dy}}{{dx}}=2$
(c) $ \displaystyle 4x^3$
Solution
(a) $ \displaystyle \int{{{{x}^{2}}dx}}=\displaystyle \frac{{{{x}^{{2+1}}}}}{{2+1}}+C$$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{3}{{x}^{3}}+C$
(b) $ \displaystyle \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=2$
$ \displaystyle \begin{array}{l}\therefore \ \ dy=2dx\\\\\therefore \ \ \displaystyle \int{{1dy}}=\displaystyle \int{{2dx}}\\\\\therefore \ \ y=2\times \displaystyle \frac{{{{x}^{{0+1}}}}}{1}+C\\\\\therefore \ \ y=2x+C\end{array}$
(c) $ \displaystyle \int{{4{{x}^{3}}dx}}=4\int{{{{x}^{3}}dx}}$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\times \displaystyle \frac{{{{x}^{{3+1}}}}}{{3+1}}+C\\\ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\left( {\displaystyle \frac{1}{4}} \right){{x}^{4}}+C\\\ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{4}}+C\end{array}$
Example (2) Find the integral of each of the following.
(a) $\displaystyle \frac{1}{x^4} $
(b) $ \displaystyle \sqrt[3]{x}$
(c) $ \displaystyle \frac{2}{\sqrt{x}}$
Solution
(a) $ \displaystyle \int{{\frac{1}{{{{x}^{4}}}}\ }}dx=\displaystyle \int{{{{x}^{{-4}}}\ }}dx$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{x}^{{-4+1}}}}}{{-4+1}}+C\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\displaystyle \frac{1}{{3{{x}^{3}}}}+C\end{array}$
(b) $ \displaystyle \int{{\sqrt[3]{x}\ }}dx=\int{{{{x}^{{\frac{1}{3}}}}\ }}dx$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{{{{x}^{{ \frac{1}{3}+1}}}}}{{\displaystyle \frac{1}{3}+1}}+C\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{3}{4}{{x}^{{\frac{4}{3}}}}+C\end{array}$
(c) $ \displaystyle \int{{\frac{2}{{\sqrt{x}}}\ }}dx=\int{{2{{x}^{{-\frac{1}{2}}}}\ }}dx$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\times \displaystyle \frac{{{{x}^{{-\frac{1}{2}+1}}}}}{{-\displaystyle \frac{1}{2}+1}}+C\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4\sqrt{x}+C\end{array}$
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