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Одређени интеграл 2

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

Задаци

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Пр.3)   Решити   $\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} - \sin x} \right)dx} $

Пр.4)   $\int\limits_{ - 2}^{ - 1} {\frac{{dx}}{{{{\left( {11 + 5x} \right)}^3}}}} $

Пр.5)   $\int\limits_2^{ - 13} {\frac{{dx}}{{\sqrt[5]{{3 - {x^4}}}}}} $

Пр.3)   $\int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} - \sin x} \right)dx}  = \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\frac{1}{{{{\cos }^2}x}}dx}  - \int\limits_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {\sin xdx}  = \left. {tgx} \right|_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} + \left. {\cos x} \right|_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} = $

$ = \left( {tgx + \left. {\cos x} \right|} \right)_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} = tg\frac{\pi }{4} + \cos \frac{\pi }{4} - \left( {tg\frac{\pi }{4} + \cos \frac{\pi }{4}} \right) = $

$ = 1 + \frac{{\sqrt 2 }}{2} - \left( { - 1 + \frac{{\sqrt 2 }}{2}} \right) = 2$

 

Пр.4)   $\int\limits_{ - 2}^{ - 1} {\frac{{dx}}{{{{\left( {11 + 5x} \right)}^3}}}} $

\[\begin{array}{*{20}{c}}
\begin{gathered}
11 + 5x = t \hfill \\
5dx = dt \hfill \\
dx = \frac{{dt}}{5} \hfill \\
\end{gathered} &\begin{gathered}
x = - 2 \hfill \\
x = - 1 \hfill \\
\end{gathered} &\begin{gathered}
t = 11 + 5\left( { - 2} \right) = 1 \hfill \\
t = 11 + 5\left( { - 1} \right) = 6 \hfill \\
\end{gathered}
\end{array}\]

$ = \int\limits_1^6 {\frac{{\frac{{dt}}{5}}}{{{t^3}}}}  = \frac{1}{5}\int\limits_1^6 {{t^{ - 3}}dt = } \frac{1}{5} \cdot \left. {\frac{{{t^{ - 2}}}}{{ - 2}}} \right|_1^6 = \frac{{ - 1}}{{10}} \cdot \left. {\frac{1}{{{t^2}}}} \right|_1^6 =  - \frac{1}{{10}} \cdot \left( {\frac{1}{{{6^2}}} - \frac{1}{{{1^2}}}} \right) =  - \frac{1}{{10}} \cdot \left( { - \frac{{35}}{{36}}} \right) = \frac{7}{{72}}$

 

Пр.5)   $\int\limits_2^{ - 13} {\frac{{dx}}{{\sqrt[5]{{3 - {x^4}}}}}} $

\[\begin{array}{*{20}{c}}
\begin{gathered}
3 - x = t \hfill \\
- dx = dt \hfill \\
dx = - dt \hfill \\
\end{gathered} &\begin{gathered}
x = 2 \hfill \\
x = - 13 \hfill \\
\end{gathered} &\begin{gathered}
t = 1 \hfill \\
t = 16 \hfill \\
\end{gathered}
\end{array}\]

$ = \int\limits_1^{16} {\frac{{ - dt}}{{{{\sqrt[5]{t}}^4}}}}  =  - \int\limits_1^{16} {{t^{ - \frac{4}{3}}}} dt =  - \left. {\frac{{{t^{\frac{1}{5}}}}}{{\frac{1}{5}}}} \right|_1^{16} =  - 5\left. {\sqrt[5]{t}} \right|_1^{16} =  - 5\sqrt[5]{{16}} - \left( { - 5\sqrt[5]{1}} \right) =  - 5\left( {\sqrt[5]{{16}} - 1} \right)$