Integration of $ \displaystyle \frac{1}{x}$, $ \displaystyle \frac{1}{ax+b}$ and exponential function

$\displaystyle \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\displaystyle \frac{d}{{dx}}\left( {\ln x} \right)=\frac{1}{x}} & {\displaystyle \int{{\frac{1}{x}}}\ dx=\ln \left| x \right|+C} \\ \hline {\displaystyle \frac{d}{{dx}}\left[ {\ln \left( {ax+b} \right)} \right]=\frac{a}{{ax+b}}} & {\displaystyle \int{{\frac{1}{{ax+b}}}}\ dx=\frac{1}{a}\ln \left| {ax+b} \right|+C} \\ \hline { \displaystyle \frac{d}{{dx}}\left( {{{e}^{x}}} \right)={{e}^{x}}} & {\displaystyle \int{{{{e}^{x}}}}\ dx={{e}^{x}}+C}\\ \hline {\displaystyle \frac{d}{{dx}}\left( {{{e}^{{ax+b}}}} \right)=a{{e}^{{ax+b}}}} & {\displaystyle \int{{{{e}^{{ax+b}}}}}\ dx=\frac{1}{a}{{e}^{{ax+b}}}+C} \\ \hline \end{array}\end{array}$

1.  Integrate each of the following with respect to $ \displaystyle x$

$ \displaystyle \begin{array}{l}\text{(a)}\ \ \displaystyle \frac{1}{{3x}}\\\\\text{(b)}\ \ \displaystyle \frac{2}{{5x}}\\\\\text{(c)}\ \ \displaystyle \frac{3}{{4x-1}}\\\\\text{(d)}\ \ \displaystyle \frac{{10}}{{25x+3}}\\\\\text{(e)}\ \ \displaystyle \frac{7}{{2-5x}}\end{array}$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \ \displaystyle \int{{\displaystyle \frac{1}{{3x}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{3}\displaystyle \int{{\displaystyle \frac{1}{x}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{3}\ln |x|+C\\\\\\\\\text{(b)}\ \ \ \ \displaystyle \int{{\displaystyle \frac{2}{{5x}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{5}\displaystyle \int{{\displaystyle \frac{1}{x}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{5}\ln |x|+C\\\\\\\\\text{(c)}\ \ \ \ \displaystyle \int{{\displaystyle \frac{3}{{4x-1}}}}\ dx\\\\\ \ \ \ =\ \ 3\displaystyle \int{{\displaystyle \frac{1}{{4x-1}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{3}{4}\ln |4x-1|+C\\\\\\\\\text{(d)}\ \ \ \ \displaystyle \int{{\displaystyle \frac{{10}}{{25x+3}}}}\ dx\\\\\ \ \ \ =\ \ 10\displaystyle \int{{\displaystyle \frac{1}{{25x+3}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{{10}}{{25}}\ln |25x+3|+C\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{5}\ln |25x+3|+C\\\\\\\\\text{(e)}\ \ \ \ \displaystyle \int{{\displaystyle \frac{7}{{2-5x}}}}\ dx\\\\\ \ \ \ =\ \ 7\displaystyle \int{{\displaystyle \frac{1}{{2-5x}}}}\ dx\\\\\ \ \ \ =\ \ -\displaystyle \frac{7}{5}\ln |2-5x|+C\end{array}$

2.  Find each of the following indefinite integrals.

$ \displaystyle \begin{array}{l}\text{(a)}\ \ \displaystyle \int{{{{e}^{{3x}}}}}\ dx\\\\\text{(b)}\ \ \displaystyle \int{{{{e}^{{2-5x}}}}}\ dx\\\\\text{(c)}\ \ \displaystyle \int{{{{e}^{{3x+\pi }}}}}\ dx\\\\\text{(d)}\ \ \displaystyle \int{{{{{\left( {{{e}^{{2x}}}+1} \right)}}^{2}}}}\ dx\end{array}$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \displaystyle \int{{{{e}^{{3x}}}}}\ dx\\\\\ \ \ \ =\displaystyle \frac{1}{3}{{e}^{{3x}}}+C\\\\\\\\\text{(b)}\ \ \ \displaystyle \int{{{{e}^{{2-5x}}}}}\ dx\\\\\ \ \ \ =-\displaystyle \frac{1}{5}{{e}^{{2-5x}}}+C\\\\\\\\\text{(c)}\ \ \ \displaystyle \int{{{{e}^{{3x+\pi }}}}}\ dx\\\\\ \ \ \ =\displaystyle \frac{1}{3}{{e}^{{3x+\pi }}}+C\\\\\\\\\text{(d)}\ \ \ \displaystyle \int{{{{{\left( {{{e}^{{2x}}}+1} \right)}}^{2}}}}\ dx\\\\\ \ \ \ =\ \displaystyle \int{{\left( {{{e}^{{4x}}}+2{{e}^{{2x}}}+1} \right)}}\ dx\\\\\ \ \ \ =\ \displaystyle \int{{{{e}^{{4x}}}}}\ dx+2\displaystyle \int{{{{e}^{{2x}}}}}\ dx+\displaystyle \int{1}\ dx\\\\\ \ \ \ =\ \displaystyle \frac{1}{4}{{e}^{{4x}}}+{{e}^{{2x}}}+x+C\end{array}$

Practice Problems for Indefinite Integration

Integration of $ \displaystyle x^n$

$\displaystyle \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\displaystyle \int{{{{x}^{n}}\ dx=\frac{{{{x}^{{n+1}}}}}{{n+1}}+C}},n\ne -1 } \\ \hline \end{array}\end{array}$

➊       Evaluate each of the following indefinite integrals.

(a) $ \displaystyle \int{{\left( {5{{x}^{3}}-10{{x}^{2}}+4} \right)}}\ dx$

(b) $ \displaystyle \int{{\left( {{{x}^{8}}+\frac{1}{{{{x}^{8}}}}} \right)}\ }dx$

(c) $ \displaystyle \int{{\left( {3\sqrt[4]{{{{x}^{3}}}}-\frac{2}{{{{x}^{5}}}}+\frac{1}{{4\sqrt{x}}}} \right)}\ }dx$

(d) $ \displaystyle \int{{\left( {x-2} \right)(x+2)\ }}dx$

(e) $ \displaystyle \int{{2x(x+3)\ }}dx$

(f) $ \displaystyle \int{{\left( {\frac{{4{{x}^{3}}-2{{x}^{2}}+3x}}{{2x}}} \right)\ }}dx$

(g) $ \displaystyle \int{{\left( {\frac{{2x-3}}{{\sqrt[3]{x}}}} \right)\ }}dx$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \ \displaystyle \int{{(5{{x}^{3}}-10{{x}^{2}}+4)}}~dx\\\\\ \ \ \ =\ \ \displaystyle \int{{5{{x}^{3}}}}~dx-\displaystyle \int{{10{{x}^{2}}\ }}dx+\displaystyle \int{4}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{5}{4}{{x}^{4}}-\displaystyle \frac{{10}}{3}{{x}^{3}}+4x+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(b)}\ \ \ \ \displaystyle \int{{\left( {{{x}^{8}}+\displaystyle \frac{1}{{{{x}^{8}}}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{{{x}^{8}}}}~dx+\displaystyle \int{{\displaystyle \frac{1}{{{{x}^{8}}}}\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{{{x}^{8}}}}~dx+\displaystyle \int{{{{x}^{{-8}}}\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{9}{{x}^{9}}-\displaystyle \frac{1}{7}{{x}^{{-7}}}+C\\\\\ \ \ \ =\ \ \displaystyle \frac{{{{x}^{9}}}}{9}-\displaystyle \frac{1}{{7{{x}^{7}}}}+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(c)}\ \ \ \ \displaystyle \int{{\left( {3\sqrt[4]{{{{x}^{3}}}}-\displaystyle \frac{2}{{{{x}^{5}}}}+\displaystyle {1}{{4\sqrt{x}}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{3\sqrt[4]{{{{x}^{3}}}}}}~dx-\displaystyle \int{{\displaystyle \frac{2}{{{{x}^{5}}}}\ }}dx+\displaystyle \int{{\displaystyle \frac{1}{{4\sqrt{x}}}\ }}dx\\\\\ \ \ \ =\ \ 3\displaystyle \int{{{{x}^{{\frac{3}{4}}}}}}~dx-2\displaystyle \int{{{{x}^{{-5}}}\ }}dx+\displaystyle \frac{1}{4}\displaystyle \int{{{{x}^{{-\frac{1}{2}}}}}}~dx\\\\\ \ \ \ =\ \ \displaystyle \frac{{12}}{7}{{x}^{{\frac{7}{4}}}}+\displaystyle \frac{1}{2}{{x}^{{-4}}}+\displaystyle \frac{1}{2}{{x}^{{\frac{1}{2}}}}+C\\\\\ \ \ \ =\ \ \displaystyle \frac{{12{{x}^{{\frac{7}{4}}}}}}{7}+\displaystyle \frac{1}{{2{{x}^{4}}}}+\displaystyle \frac{{\sqrt{x}}}{2}+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(d)}\ \ \ \ \displaystyle \int{{\left( {x-2} \right)\left( {x+2} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {{{x}^{2}}-4} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{{{x}^{2}}\ }}dx-\displaystyle \int{{4\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{3}{{x}^{3}}-4x+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(e)}\ \ \ \ \displaystyle \int{{2x\left( {x+3} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {2{{x}^{2}}-6x} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{2{{x}^{2}}\ }}dx-\displaystyle \int{{6x\ }}dx\\\\\ \ \ \ =\ \ 2\displaystyle \int{{{{x}^{2}}\ }}dx-6\displaystyle \int{{x\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{3}{{x}^{3}}-3{{x}^{2}}+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(f)}\ \ \ \ \displaystyle \int{{\left( {\displaystyle \frac{{4{{x}^{3}}-2{{x}^{2}}+3x}}{{2x}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {\displaystyle \frac{{4{{x}^{3}}}}{{2x}}-\displaystyle \frac{{2{{x}^{2}}}}{{2x}}+\displaystyle \frac{{3x}}{{2x}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {2{{x}^{2}}-x+\displaystyle \frac{3}{2}} \right)\ }}dx\\\\\ \ \ \ =\ \ 2\displaystyle \int{{{{x}^{2}}\ }}dx-\displaystyle \int{{x\ }}dx+\displaystyle \frac{3}{2}\displaystyle \int{{dx}}\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{3}{{x}^{3}}-\displaystyle \frac{1}{2}{{x}^{2}}+\displaystyle \frac{3}{2}x+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(g)}\ \ \ \ \displaystyle \int{{\left( {\displaystyle \frac{{2x-3}}{{\sqrt[3]{x}}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {2{{x}^{{ \frac{2}{3}}}}-3{{x}^{{-\frac{1}{3}}}}} \right)\ }}dx\\\\\ \ \ \ =\ \ 2\displaystyle \int{{{{x}^{{\frac{2}{3}}}}\ }}dx-3\displaystyle \int{{{{x}^{{-\frac{1}{3}}}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{6}{5}{{x}^{{\frac{5}{3}}}}-\displaystyle \frac{9}{2}{{x}^{{\frac{2}{3}}}}+C\end{array}$

➋       Given that $ \displaystyle \frac{{dy}}{{dx}}=3{{x}^{2}}-4x-5$ and $ \displaystyle y=-4$ when $ \displaystyle x=-2,$ find the value of $ \displaystyle y$ when $ \displaystyle x=1.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=3{{x}^{2}}-4x-5\\\\\therefore \ \ \ \ dy=\left( {3{{x}^{2}}-4x-5} \right)dx\\\\\therefore \ \ \ \ y=\displaystyle \int{{\left( {3{{x}^{2}}-4x-5} \right)}}dx\\\\\ \ \ \ \ \ y={{x}^{3}}-2{{x}^{2}}-5x+C\\\\\ \ \ \ \ \ \text{When}\ x=-2,\ y=-4\\\\\therefore \ \ \ \ {{\left( {-2} \right)}^{3}}-2{{\left( {-2} \right)}^{2}}-5\left( {-2} \right)+C=-4\\\\\therefore \ \ \ \ C=2\\\\\therefore \ \ \ \ y={{x}^{3}}-2{{x}^{2}}-5x+2\\\\\therefore \ \ \ \ \text{When}\ x=1,\ y=-4\end{array}$

➌       If $ \displaystyle {g}'(x)= 3x^2-2x-3$ and $ \displaystyle g(-1)=4,$ factorize g(x) completely.

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$ \displaystyle \begin{array}{l}\ \ \ \ {g}'(x)=3{{x}^{2}}-2x-3\\\\\therefore \ \ g(x)=\displaystyle \int{{{g}'(x)}}\ dx\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \int{{\left( {3{{x}^{2}}-2x-3} \right)}}\ dx\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ = {{x}^{3}}-{{x}^{2}}-3x+C\\\\\ \ \ \ \text{Since}\ g(-1)=4,\\\\\ \ \ \ -1-1+3+C=4\\\\\therefore \ \ C=3\\\\\therefore \ \ g(x)={{x}^{3}}-{{x}^{2}}-3x+3\\\\\ \ \ \ g(1)=1-1-3+3=0\end{array}$ $ \displaystyle \therefore \ \ x-1$ is a factor of $ \displaystyle g(x).$ $\displaystyle \begin{matrix}\begin{array}{r} \left. 1 \right) \\ ~ \end{array} & \underline { \begin{array}{rrrr} 1&-1&-3&3\\ ~~~~&1&0&-3 \end{array} } \\ ~ & \begin{array}{rrrr} \ \ { 1 }& \ \ \ { 0 }& { -3 }&\ \ \color{red}{ 0 } \end{array}\end{matrix}$ $ \displaystyle \begin{array}{l}\therefore \ \ g(x)=(x-1)({{x}^{2}}-3)\\\\\therefore \ \ g(x)=(x-1)(x-\sqrt{3})(x+\sqrt{3})\end{array}$

➍       Given that the curve which has gradient $ \displaystyle \frac{dy}{dx}=2x^2 + 7x$ at any point on the curve passes through the origin, determine the equation of the curve.

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$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \frac{{dy}}{{dx}}=2{{x}^{2}}+7x\\\\\therefore \ \ dy=\left( {2{{x}^{2}}+7x} \right)dx\\\\\therefore \ \ y=\displaystyle \int{{\left( {2{{x}^{2}}+7x} \right)}}\ dx\\\\\therefore \ \ y=\displaystyle \frac{2}{3}{{x}^{3}}+\displaystyle \frac{7}{2}{{x}^{2}}+C\end{array}$ Since the curve passes through the origin, when $ \displaystyle x=0$ and $ \displaystyle y=0$. $ \displaystyle \begin{array}{l}\therefore \ \ 0=\displaystyle \frac{2}{3}\left( 0 \right)+\displaystyle \frac{7}{2}\left( 0 \right)+C\\\\\therefore \ \ C=0\\\\\therefore \ \ y=\displaystyle \frac{2}{3}{{x}^{3}}+\displaystyle \frac{7}{2}{{x}^{2}}\end{array}$

➎       The rate of change of $ \displaystyle A$ with respect to $ \displaystyle r$ is given by $ \displaystyle \frac{{dA}}{{dr}}=6r+1$. If $ \displaystyle A = 3$ when $ \displaystyle r = 1$, find $ \displaystyle A$ in terms of $ \displaystyle r$.

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$ \displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{{dA}}{{dr}}=6r+1\\\\\therefore \ dA=\left( {6r+1} \right)dr\\\\\therefore \ A=\displaystyle \int{{\left( {6r+1} \right)}}\ dr\\\\\therefore \ A=3{{r}^{2}}+r+C\end{array}$ When $ \displaystyle r=1$ and $ \displaystyle A=3$. $ \displaystyle \begin{array}{l}\therefore \ 3=3{{\left( 1 \right)}^{2}}+\left( 1 \right)+C\\\\\therefore \ C=-1\\\\\therefore \ A=3{{r}^{2}}+r-1\end{array}$

➏       If the curve which has gradient $ \displaystyle \frac{dy}{dx}=kx-1$ at any point on the curve cuts the $ \displaystyle x$-axis at $ \displaystyle x=-3$ and $ \displaystyle x=4$, determine the value of $ \displaystyle k$ and hence find the equation of the curve.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=kx-1\\\\\therefore \ \ \ dy=\left( {kx-1} \right)dx\\\\\therefore \ \ \ y=\displaystyle \int{{\left( {kx-1} \right)}}\ dx\\\\\ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{k}{2}{{x}^{2}}-x+C\end{array}$ Since the curve cuts the $ \displaystyle x$-axis at $ \displaystyle x=-3$ and $ \displaystyle x=4$ $ \displaystyle \begin{array}{l}\ \ \ \ \ \displaystyle \frac{k}{2}{{\left( {-3} \right)}^{2}}-\left( {-3} \right)+C=0\\\\\therefore \ \ \ 9k+6+2C=0\ ---(1)\\\\\ \ \ \ \ \displaystyle \frac{k}{2}{{\left( 4 \right)}^{2}}-\left( 4 \right)+C=0\\\\\therefore \ \ \ 16k-8+2C=0\ ---(2)\\\\\ \ \ \ \ \text{By}\ (2)-(1),\\\\\ \ \ \ \ 7k-14=0\\\\\therefore \ \ \ k=2\\\\\therefore \ \ \ 9(2)+6+2C=0\ \\\\\therefore \ \ \ C=-12\\\\\therefore \ \ \ y={{x}^{2}}-x-12\ \ \end{array}$

➐       If $ \displaystyle f''\left( x \right) = 15\sqrt x + 5{x^3} + 6$, $ \displaystyle f\left( 1 \right) = - \frac{5}{4}$ and $ \displaystyle f\left( 4 \right) = 404$, find the expression for $ \displaystyle f(x)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {f}''\left( x \right)=15\sqrt{x}+5{{x}^{3}}+6\\\\\therefore \ \ {f}'\left( x \right)=\displaystyle \int{{{f}''\left( x \right)}}\ dx\\\\\therefore \ \ {f}'\left( x \right)=\displaystyle \int{{\left( {15\sqrt{x}+5{{x}^{3}}+6} \right)}}\ dx\\\\\therefore \ \ {f}'\left( x \right)=10{{x}^{{\displaystyle \frac{3}{2}}}}+\displaystyle \frac{5}{4}{{x}^{4}}+6x+c\\\\\therefore \ \ f\left( x \right)=\displaystyle \int{{{f}'\left( x \right)}}\ dx\\\\\therefore \ \ f\left( x \right)=\displaystyle \int{{\left( {10{{x}^{{ \frac{3}{2}}}}+\displaystyle \frac{5}{4}{{x}^{4}}+6x+c} \right)}}\ dx\\\\\therefore \ \ f\left( x \right)=4{{x}^{{ \frac{5}{2}}}}+\displaystyle \frac{1}{4}{{x}^{5}}+3{{x}^{2}}+cx+d\\\\\ \ \ \ f(1)=-\displaystyle \frac{5}{4}\ \ \ \left[ {\text{given}} \right]\\\\\therefore \ \ 4+\displaystyle \frac{1}{4}+3+c+d=-\displaystyle \frac{5}{4}\\\\\therefore \ \ c+d=-\displaystyle \frac{{17}}{2}\ \ ---(1)\\\\\ \ \ \ f(4)=404\ \ \ \left[ {\text{given}} \right]\\\\\therefore \ \ 128+231+48+4c+d=404\\\\\therefore \ \ 4c+d=-28\ \ ---(2)\\\\\therefore \ \ c=-\displaystyle \frac{{13}}{2}\ \operatorname{and}\ d=-2\\\\\therefore \ \ f\left( x \right)=4{{x}^{{\frac{5}{2}}}}+\displaystyle \frac{1}{4}{{x}^{5}}+3{{x}^{2}}-\displaystyle \frac{{13}}{2}x-2\end{array}$

➑       Given that a ball moves along a grove with the velocity $ \displaystyle v=\frac{{ds}}{{dt}}=32t-2$ ft/s and at the time $ \displaystyle t=\frac{1}{2}\ \text{s}$ the ball is $ \displaystyle 4\ \text{ft}$ away from the starting point. Express the position of the ball as a function of time.

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$ \displaystyle \begin{array}{l}\ \ \ \ v=\displaystyle \frac{{ds}}{{dt}}=32t-2\\\\\therefore \ \ ds=\left( {32t-2} \right)dt\\\\\therefore \ \ s=\displaystyle \int{{\left( {32t-2} \right)}}\ dt\\\\\therefore \ \ s=16{{t}^{2}}-2t+C\\\\\ \ \ \ \text{When}\ t=\displaystyle \frac{1}{2},\ s=4\\\\\therefore \ \ 16{{\left( {\displaystyle \frac{1}{2}} \right)}^{2}}-2\left( {\displaystyle \frac{1}{2}} \right)+C=4\\\\\therefore \ \ C=1\\\\\therefore \ \ s=16{{t}^{2}}-2t+1\end{array}$

Integration of $ \displaystyle (ax + b)^n$

$\displaystyle \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\displaystyle \int{{{{{\left( {ax+b} \right)}}^{n}}}}\ dx=\displaystyle \frac{{{{{\left( {ax+b} \right)}}^{{n+1}}}}}{{a\left( {n+1} \right)}}+C,} \\ {\text{where}\ n\ne -1,\ a\ne 0 \ \ \ \ \ \ \ \ \ }\\ \hline \end{array}\end{array}$

➒       Integrate each of the following with respect to $ \displaystyle x$.

(a) $ \displaystyle (2x+1)^6$

(b) $ \displaystyle \frac{2}{{\sqrt{{3x-2}}}}$

(c) $ \displaystyle (5x-11)^{10}$

(d) $ \displaystyle {{\left( {\frac{3}{{4x-1}}} \right)}^{3}}$

(e) $ \displaystyle \frac{2}{{5{{{\left( {3x-1} \right)}}^{4}}}}$

(f) $ \displaystyle \sqrt{{{{{\left( {5x+2} \right)}}^{3}}}}$

(g) $ \displaystyle \frac{5}{{2\sqrt{{{{{\left( {3x-1} \right)}}^{7}}}}}}$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \ \ \displaystyle \int{{{{{(2x+1)}}^{6}}}}\ dx\ \\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{7\times 2}}{{(2x+1)}^{7}}+C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{14}}{{(2x+1)}^{7}}+C\\\\\\\\\text{(b)}\ \ \ \ \ \displaystyle \int{{\displaystyle \frac{2}{{\sqrt{{3x-2}}}}}}\ dx\\\\\ \ \ \ \ =\ \ 2\displaystyle \int{{{{{\left( {3x-2} \right)}}^{{- \frac{1}{2}}}}}}\ dx\\\\\ \ \ \ \ =\ \ 2\times \displaystyle \frac{{{{{\left( {3x-2} \right)}}^{{\frac{1}{2}}}}}}{{\displaystyle \frac{1}{2}\times 3}}\ +C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{4}{3}\sqrt{{3x-2}}+C\\\\\\\\\text{(c)}\ \ \ \ \ \displaystyle \int{{{{{(5x-11)}}^{{10}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{11\times 5}}{{(5x-11)}^{{10}}}+C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{55}}{{(5x-11)}^{{10}}}+C\\\\\\\\\text{(d)}\ \ \ \ \ \displaystyle \int{{{{{\left( {\displaystyle \frac{3}{{4x-1}}} \right)}}^{3}}}}\ dx\\\\\ \ \ \ \ =\ \ 27\displaystyle \int{{{{{\left( {4x-1} \right)}}^{{-3}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{{27}}{{-2\times 4}}{{\left( {4x-1} \right)}^{{-2}}}+C\\\\\ \ \ \ \ =\ \ -\displaystyle \frac{{27}}{{8{{{\left( {4x-1} \right)}}^{2}}}}+C\\\\\\\\\text{(e)}\ \ \ \ \ \displaystyle \int{{\displaystyle \frac{2}{{5{{{\left( {3x-1} \right)}}^{4}}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{2}{5}\displaystyle \int{{{{{\left( {3x-1} \right)}}^{{-4}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{2}{{5(-3)(3)}}{{\left( {3x-1} \right)}^{{-3}}}+C\\\\\ \ \ \ \ =\ \ -\displaystyle \frac{2}{{45{{{\left( {3x-1} \right)}}^{3}}}}+C\\\\\\\\\text{(f)}\ \ \ \ \ \displaystyle \int{{\sqrt{{{{{\left( {5x+2} \right)}}^{3}}}}}}\ dx\\\\\ \ \ \ \ =\ \displaystyle \int{{{{{\left( {5x+1} \right)}}^{{\frac{3}{2}}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{\displaystyle \frac{5}{2}\times 5}}{{\left( {5x+1} \right)}^{{\displaystyle \frac{5}{2}}}}+C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{2}{{25}}{{\left( {5x+1} \right)}^{{\frac{5}{2}}}}+C\\\\\\\\\text{(g)}\ \ \ \ \ \displaystyle \int{{\displaystyle \frac{5}{{2\sqrt{{{{{\left( {3x-1} \right)}}^{7}}}}}}}}\ dx\\\\\ \ \ \ \ =\ \displaystyle \frac{5}{2}\displaystyle \int{{{{{\left( {3x-1} \right)}}^{{-\frac{7}{2}}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{5}{2}{{\left( {3x-1} \right)}^{{- \frac{5}{2}}}}+C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{5}{{2\left( {-\displaystyle \frac{5}{2}} \right)3}}{{\left( {3x-1} \right)}^{{- \frac{5}{2}}}}+C\\\\\ \ \ \ \ =\ \ -\displaystyle \frac{1}{{3{{{\left( {3x-1} \right)}}^{{ \frac{5}{2}}}}}}+C\end{array}$

➓       Find the equation of the curve which cuts the $ \displaystyle y$-axis at $ \displaystyle (0, 4)$ and for which $ \displaystyle \frac{{dy}}{{dx}}={{(2x+1)}^{3}}$.

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$ \displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{{dy}}{{dx}}={{(2x+1)}^{3}}\\\\\therefore \ \ dy={{(2x+1)}^{3}}dx\\\\\therefore \ \ y=\displaystyle \int{{{{{(2x+1)}}^{3}}}}\ dx\\\\\therefore \ \ y=\displaystyle \frac{1}{{4\times 2}}{{(2x+1)}^{4}}+C\\\\\therefore \ \ y=\displaystyle \frac{1}{8}{{(2x+1)}^{4}}+C\end{array}$ Since the curve passes through $ \displaystyle (0, 4)$, $ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \frac{1}{8}{{(2(0)-1)}^{4}}+C=4\\\\\therefore \ \ \ C=\displaystyle \frac{{31}}{8}\\\\\therefore \ \ y=\displaystyle \frac{1}{8}{{(2x+1)}^{4}}+\displaystyle \frac{{31}}{8}\\\\\therefore \ \ y=\displaystyle \frac{1}{8}\left[ {{{{(2x+1)}}^{4}}+31} \right]\end{array}$

Indefinite Integral (Anti-Derivative)

 

$ \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.

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}}$ $\displaystyle y={{x}^{3}}+c\Rightarrow \frac{{dy}}{{dx}}=3{{x}^{2}},c =\text{constant}$

ဒါကြောင့် ...

$ \displaystyle \frac{{dy}}{{dx}}=3{{x}^{2}}\Rightarrow y=\int{{3{{x}^{2}}dx=}}{{x}^{3}}+c$

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$

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}$

Area under a Curve (Introduction to Definite Integral)

ပုံသဏ္ဌာန် မှန်သော geometry ရုပ်ပုံ တစ်ခုရဲ့  ဧရိယာကို အလယ်တန်း အဆင့်မှာကတည်း ရှာတတ်ခဲ့ မှာပါ။ ဥပမာ ...

Shape	Diagram	Area Formula
Circle		$ \displaystyle A=\pi r^2$
Triangle		$ \displaystyle A=\frac{1}{2}bh$
Rectangle		$ \displaystyle A=lw$
Polygon		$ \displaystyle A=A_1+A_2+A_3+A_4$

သို့သော် ပုံသဏ္ဌာာန် မတိကျသော မျဉ်းကွေးတစ်ခု ၏ အောက်ရှိ အစိပ်အပိုင်း တစ်ခု၏ ဧရိယာကိုတော့ အထက်ပါ အတိုင်းပုံသေနည်း ထုတ်၍ ရှာရန် မလွယ်ကူတော့ပါ။ အောက်ပါပုံကို ကြည့်ပါ။



အထက်ပါ ပုံကို ကြည့်လျှင် $ \displaystyle x=0$ မှ $ \displaystyle x=1$ အတွင်း $ \displaystyle y=x^2$ ဆိုသော မျဉ်းကွေးအောက်ရှိ ဧရိယာ ( $ \displaystyle S$) ကို တိကျစွာရှာရန်အတွက် သတ်မှတ်ထားသော ပုံသေနည်းကို အလွယ်တကူ မရနိုင်တော့ပေ။ ထိုဧရိယာကို ရှာရန် နည်လမ်းကို အောက်ပါ အတိုင်း စဉ်းစားကြည့်မည်။


ပုံ တွင်မြင်တွေ့ရသည့် အတိုင်း လိုချင်သော ဧရိယာကို  $ \displaystyle S_1, S_2, S_3, S_4$ ဟူ၍ အပိုင်းလေးပိုင်း ပိုင်း၍ရှာပြီးမှ ပြန်ပေါင်းလျှင် ရနိုင်ပါမည်။ သို့သော်လည်း ထိုတစ်ပိုင်းစီကို မည်ကဲ့သို့ ရှာမည်နည်း။ တိကျသော ပုံသေနည်း မရှိသေး။ ထို့ကြောင့် ပုံသေနည်း ရရန်အောက်ပါ အတိုင်း ပြုပြင်ပြီး စဉ်းစားကြည့်မည်။

အပိုင်း တစ်ပိုင်းစီ၏ လက်ယာဘက်အစွန်းမှ ထောင့်မှန်စတုဂံများ တည်ဆောက်ပြီး ၎င်း  ထောင့်မှန်စတုဂံ တစ်ခုစီ၏ ဧရိယာများကို ရှာ၍ ပြန်ပေါင်းလျှင် လိုချင်သော ဧရိယာကို အနီးစပ်ဆုံး ရနိုင်မည်ဖြစ်သည်။

ထောင့်မှန်စတုဂံ၏ ဧရိယာ	အခြေ × အမြင့်
ပထမထောင့်မှန်စတုဂံ၏ ဧရိယာ	$ \displaystyle S_1=\frac{1}{4}\times f\left( {\frac{1}{4}} \right)=\frac{1}{4}\times \frac{1}{{16}}=\frac{1}{{64}}$
ဒုတိယထောင့်မှန်စတုဂံ၏ ဧရိယာ	$ \displaystyle S_2=\frac{1}{4}\times f\left( {\frac{1}{2}} \right)=\frac{1}{4}\times \frac{1}{4}=\frac{1}{{16}}=\frac{4}{{64}}$
တတိယထောင့်မှန်စတုဂံ၏ ဧရိယာ	$ \displaystyle S_3=\frac{1}{4}\times f\left( {\frac{3}{4}} \right)=\frac{1}{4}\times \frac{9}{{16}}=\frac{9}{{64}}$
စတုတ္ထထောင့်မှန်စတုဂံ၏ ဧရိယာ	$ \displaystyle S_4=\frac{1}{4}\times f\left( 1 \right)=\frac{1}{4}\times 1=\frac{1}{4}=\frac{{16}}{{64}}$

ထို့ကြောင့် ထောင့်မှန်စတုဂံ အားလုံး၏ ဧရိယာ = $ \displaystyle \frac{{30}}{{64}}=0.46875$ ဖြစ်မည်။ သို့ရာတွင် လိုချင်သော ဧရိယာသည် ထောင့်မှန်စတုဂံ အားလုံး၏ ဧရိယာ အောက်ငယ်နေသည် ကို ပုံတွင်အထင်အရှား တွေ့နိုင်ပေသည်။ ထို့ကြောင့် နောက်ထပ် တစ်နည်း စဉ်းစားကြည့်မည်။

ယခုတစ်ခါ အပိုင်း တစ်ပိုင်းစီ၏ လက်ဝဲဘက်အစွန်းမှ ထောင့်မှန်စတုဂံများ တည်ဆောက်ပြီး ၎င်း  ထောင့်မှန်စတုဂံ တစ်ခုစီ၏ ဧရိယာများကို ရှာ၍ ပြန်ပေါင်း ကြည့်မည်။

ထောင့်မှန်စတုဂံ၏ ဧရိယာ	အခြေ × အမြင့်
ပထမထောင့်မှန်စတုဂံ၏ ဧရိယာ	$ \displaystyle \frac{1}{4}\times f\left( 0 \right)=\frac{1}{4}\times 0=0$
ဒုတိယထောင့်မှန်စတုဂံ၏ ဧရိယာ	$ \displaystyle \frac{1}{4}\times f\left( {\frac{1}{4}} \right)=\frac{1}{4}\times \frac{1}{{16}}=\frac{1}{{64}}$
တတိယထောင့်မှန်စတုဂံ၏ ဧရိယာ	$ \displaystyle \frac{1}{4}\times f\left( {\frac{1}{2}} \right)=\frac{1}{4}\times \frac{1}{4}=\frac{1}{{16}}=\frac{4}{{64}}$
စတုတ္ထထောင့်မှန်စတုဂံ၏ ဧရိယာ	$ \displaystyle \frac{1}{4}\times f\left( {\frac{3}{4}} \right)=\frac{1}{4}\times \frac{9}{{16}}=\frac{9}{{64}}$

ထို့ကြောင့် ထောင့်မှန်စတုဂံ အားလုံး၏ ဧရိယာ = $ \displaystyle \frac{{14}}{{64}}=0.21875$ ဖြစ်မည်။ သို့ရာတွင် လိုချင်သော ဧရိယာသည် ထောင့်မှန်စတုဂံ အားလုံး၏ ဧရိယာ ထက်ကြီးနေသည် ကို ပုံတွင်အထင်အရှား တွေ့နိုင် ပြန်ပါသည်။

ထို့ကြောင့် လိုချင်သော ဧရိယာ $ \displaystyle S$ သည်...

$ \displaystyle 0.21875 < S < 0.46875$ ဖြစ်ပေမည်။ တိကျသော အဖြေကို မရသေးပါ။

ပိုမိုတိကျသော အဖြေရနိုင်ရန် အထက်ပါနည်းအတိုင်း ထောင့်မှန်စတုဂံ အရေအတွက်ကို တိုး၍ စိပ်ပိုင်းလိုက် သည့်အခါ ...


$ \displaystyle 0.2734375 < S < 0.3984375$ ဖြစ်လာပြန်ပါသည်။ ပုံကိုကြည့်ခြင်းအား ဖြင့် $ \displaystyle S$ ၏ တန်ဖိုးသည် အဖြေမှန်နှင့် ပိုမို နီးစပ်လာသည်ကို တွေ့ရပေမည်။ ဤ ဥပမာများကို ကြည့်လျှင် စိပ်ပိုင်းလိုက်သည့် ထောင့်မှန်စတုဂံ အရေအတွက် ပို၍များလာလေလေ အဖြေမှန်နှင့် ပို၍ နီးစပ်လာသည်ကို တွေ့ရပေမည်။

$ \displaystyle n$	$ \displaystyle L_n$	$ \displaystyle R_n$
10	0.2850000	0.3850000
20	0.3087500	0.3587500
30	0.3168519	0.3501852
50	0.3234000	0.3434000
100	0.3283500	0.3383500
1000	0.3328335	0.3338335

အထက်ပါ ထောင့်မှန်စတုဂံများ၏ ဧရိယာများပေါင်းလဒ်ကို Riemann sum ဟုခေါ်သည်။

အောက်ပါ geogebra applet ဖြင့်လေ့လာကြည့်ပါ။


applet တွင် တွေ့ရှိချက်အရ left sum, right sum တို့ထက် midpoint sum သည် အဖြေမှန်နှင့် ပိုမို နီးစပ်သည်ကို တွေ့ရှိရပေမည်။ သို့ရာတွင် စိပ်ပိုင်းသည့် အရေအတွက် အနန္တ (∞) သို့ ချဉ်းကပ်သွား သည့်အခါ left sum နှင့် right sum တို့သည့်ကွားခြားမှု မရှိတော့ပေ။
 
$ \displaystyle x = a$ နှင့်  $ \displaystyle x=b$ ကြာား ပေးထားသော curve ၏ အောက်ရှိ လိုချင်သော ဧရိယာကို ထောင့်မှန်စတုဂံပေါင်း $ \displaystyle n$ ခု စိပ်ပိုင်းလိုက်ပြီး တစ်ခုစီ၏ အခြေအလျားသည် $ \displaystyle \Delta x$ ရှိသည်ဆိုပါစို့။

ထိုအခါ $ \displaystyle \Delta x=\frac{{b-a}}{n}$ ဖြစ်မည်။

ထောင့်မှန်စတုဂံ တစ်ခုစီ၏ ဧရိယာကို အောက်ပါအတိုင်း ရှာယူနိုင်ပါသည်။

ပထမ ထောင့်မှန်စတုဂံ၏ ဧရိယာ	$ \displaystyle f({{x}_{1}})\Delta x$
ဒုတိယ ထောင့်မှန်စတုဂံ၏ ဧရိယာ	$ \displaystyle f({{x}_{2}})\Delta x$
တတိယ ထောင့်မှန်စတုဂံ၏ ဧရိယာ	$ \displaystyle f({{x}_{3}})\Delta x$
. . .	. . .
i အကြိမ်မြောက် ထောင့်မှန်စတုဂံ၏ ဧရိယာ	$ \displaystyle f({{x}_{i}})\Delta x$

ထို့ကြောင့် ထောင့်မှန်စတုဂံပေါင်း $ \displaystyle n$ ခု စိပ်ပိုင်းထားလျှင် စုစုပေါင်း ဧရိယာသည်...

$ \displaystyle {{S}_{n}}=f({{x}_{1}})\Delta x+f({{x}_{2}})\Delta x+f({{x}_{3}})\Delta x+...+f({{x}_{n}})\Delta x$

အထက်တွင် သိပြီးဖြစ်သည့်အတိုင်း ထောင့်မှန်စတုဂံ အရေအတွက် များလာလေလေ ဧရိယာ၏ တန်ဖိုးသည်ပို၍ တိကျလေလေ ဖြစ်ရာ သတ်မှတ်ထားသော interval $ \displaystyle a$ နှင့် $ \displaystyle b$ ကြားတွင် ထောင့်မှန်စတုဂံ အရေအတွက် $ \displaystyle n$ သည် $ \displaystyle ∞$ သို့ချဉ်းကပ်သွားသည့် အခါ လိုချင်သော ဧရိယာကို အတိအကျ (exact value) ရှာယူနိုင်ပါသည်။

ထို့ကြောင့် လိုချင်သော ဧရိယာ $ \displaystyle S$ ကို အောက်ပါအတိုင်း တွက်ယူနိုင်ပါသည်။ 

$ \displaystyle S=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{f({{x}_{i}})\Delta x}}$ $ \displaystyle \ \ \ =\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left[ {f({{x}_{1}})\Delta x+f({{x}_{2}})\Delta x+f({{x}_{3}})\Delta x+...+f({{x}_{n}})\Delta x} \right]$

အထက်ပါ ဧရိယာသည် ပေးထားသော curve အောက်ရှိ သတ်မှတ်ထားသော interval $ \displaystyle a$ နှင့် $ \displaystyle b$ ကြားတွင်ရှိသောကြောင့် ဧရိယာရှာရန် ပုံသေနည်း notation ကို အောက်ပါအတိုင်း ပြောင်းရေးပါသည်။

$ \displaystyle S=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=1}}^{n}{{f({{x}_{i}})\Delta x}}=\int_{a}^{b}{{f(x)dx}}$

The Definite Integral as the Area of a Region


If $ \displaystyle f$ is continuous and nonnegative on the closed interval $ \displaystyle [a, b]$, then the area of the region bounded by the graph of $ \displaystyle f$, the $ \displaystyle x$-axis, and the vertical lines $ \displaystyle x = a$ and $ \displaystyle x = b$ is $ \displaystyle A$, then

$ \displaystyle A=\int_{a}^{b}{{f(x)dx}}$