Примена одређеног интеграла 3

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Пр.3)   Израчунај површину ограничену кривама $y = {x^2} + 4x$ и

            $y = x + 4$.

$P:y = {x^2} + 4x$

нула: $y = 0$

${x^2} + 4 = 0$

$x\left( {x + 4} \right) = 0$

\[\begin{array}{*{20}{c}}
\begin{gathered}
x = 0 \hfill \\
\hfill \\
\end{gathered} &{}&\begin{gathered}
x + 4 = 0 \hfill \\
x = - 4 \hfill \\
\end{gathered}
\end{array}\]

$p:y = x + 4$

нула: $y = 0$

$x + 4 = 0$

$x =  - 4$

\[P \cap p:\left\{ \begin{gathered}
y = {x^2} + 4x \hfill \\
y = x + 4 \hfill \\
\end{gathered} \right.\]

${x^2} + 4x = x + 4$

${x^2} + 3x - 4 = 0$

${x_{1,2}} = \frac{{ - 3 \pm \sqrt {9 + 16} }}{2} = \frac{{ - 3 \pm 5}}{2}$

\[\begin{array}{*{20}{c}}
\begin{gathered}
{x_1} = - 4 \hfill \\
{y_1} = 0 \hfill \\
A\left( { - 4;0} \right) \hfill \\
\end{gathered} &{}&\begin{gathered}
{x_2} = 1 \hfill \\
{y_2} = 5 \hfill \\
B\left( {1;5} \right) \hfill \\
\end{gathered}
\end{array}\]

$P = \int\limits_{ - 4}^1 {\left( {x + 4} \right)} dx - \int\limits_0^1 {\left( {{x^2} + 4x} \right)} dx - \int\limits_{ - 4}^0 {\left( {{x^2} + 4x} \right)} dx = $

$ = \left. {\left( {\frac{{{x^2}}}{2} + 4x} \right)} \right|_{ - 4}^1 - \left. {\left( {\frac{{{x^3}}}{3} + 4\frac{{{x^2}}}{2}} \right)} \right|_0^1 - \left. {\left( {\frac{{{x^3}}}{3} + 4\frac{{{x^2}}}{2}} \right)} \right|_{ - 4}^0 = $

$ = \frac{1}{2} + 4 - \left( {\frac{{\left( { - 4} \right)}}{2} + 4\left( { - 4} \right)} \right) - \left( {\frac{1}{3} + 2} \right) - \left( {0 - \left( {\frac{{{{\left( { - 4} \right)}^3}}}{3} + 2{{\left( { - 4} \right)}^2}} \right)} \right) = $

$ = \frac{1}{2} + 4 - 8 + 16 - \frac{1}{3} - 2 - \frac{{64}}{3} + 32 = 42 + \frac{1}{2} - \frac{{65}}{3} = \frac{{252 + 3 - 130}}{6} = \frac{{125}}{6}$