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Одређени интеграл 3

Одређени интеграл. Основна својства. Метода смене код одређеног интеграла. Сложенији примери.

Задаци

Текст задатака објашњених у видео лекцији:

Пр.6)   Решити   $\int\limits_e^{{e^2}} {\frac{{dx}}{{x\ln x}}} $

Пр.7)   $\int\limits_2^{3,5} {\frac{{dx}}{{\sqrt {5 + 4x - {x^2}} }}} $

Пр.6)   $\int\limits_e^{{e^2}} {\frac{{dx}}{{x\ln x}}}= $

\[\begin{array}{*{20}{c}}
\begin{gathered}
\ln x = t \hfill \\
\frac{1}{x}dx = dt \hfill \\
\end{gathered} &\begin{gathered}
x = e \hfill \\
x = {e^2} \hfill \\
\end{gathered} &\begin{gathered}
t = \ln e = 1 \hfill \\
t = \ln {e^2} = 2\ln e = 2 \hfill \\
\end{gathered}
\end{array}\]

$ = \int\limits_1^2 {\frac{{dt}}{t}}  = \left. {\operatorname{lnt} } \right|_1^2 = \ln 2 - \underbrace {\ln 1}_{ = 0} = \ln 2$

 

Пр.7)   $\int\limits_2^{3,5} {\frac{{dx}}{{\sqrt {5 + 4x - {x^2}} }}}  = \int\limits_2^{3,5} {\frac{{dx}}{{\sqrt {5 + 4 - 4 + 4x - {x^2}} }}}  = \int\limits_2^{3,5} {\frac{{dx}}{{\sqrt {9 - \left( {4 + 4x - {x^2}} \right)} }}}  = $

$ = \int\limits_2^{3,5} {\frac{{dx}}{{\sqrt {9 - {{\left( {2 - x} \right)}^2}} }}}  = \int\limits_2^{3,5} {\frac{{dx}}{{\sqrt {9\left( {1 - \frac{{{{\left( {2 - x} \right)}^2}}}{9}} \right)} }}}  = \int\limits_2^{3,5} {\frac{{dx}}{{3\sqrt {\left( {1 - {{\left( {\frac{{2 - x}}{3}} \right)}^2}} \right)} }}}  = $

\[\begin{array}{*{20}{c}}
\begin{gathered}
\frac{{2 - x}}{3} = t \hfill \\
- \frac{1}{3}dx = dt \hfill \\
dx = - 3dt \hfill \\
\end{gathered} &\begin{gathered}
x = 2 \hfill \\
x = 3,5 \hfill \\
\end{gathered} &\begin{gathered}
t = 0 \hfill \\
t = \frac{{2 - 2,5}}{3} = \frac{{ - 1,5}}{3} = - \frac{1}{2} \hfill \\
\end{gathered}
\end{array}\]

$ = \frac{1}{3}\int\limits_0^{ - \frac{1}{2}} {\frac{{ - 3dt}}{{\sqrt {1 - {t^2}} }}}  =  - \int\limits_0^{ - \frac{1}{2}} {\frac{{dt}}{{\sqrt {1 - {t^2}} }}}  =  - \left. {\arcsin t} \right|_0^{ - \frac{1}{2}} =  - \arcsin \left( { - \frac{1}{2}} \right) + \arcsin 0 = $

\[\begin{array}{*{20}{c}}
\begin{gathered}
\arcsin \left( { - \frac{1}{2}} \right) = x \hfill \\
- \frac{1}{2} = \sin x \hfill \\
x = - \frac{\pi }{6} \hfill \\
\end{gathered} &{}&\begin{gathered}
\arcsin \left( 0 \right) = y \hfill \\
0 = \sin y \hfill \\
y = 0 \hfill \\
\end{gathered}
\end{array}\]

$ = \frac{\pi }{6} + 0 = \frac{\pi }{6}$