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: $y = 0$

${x^2} + 4 = 0$

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

$$\begin{array}{cccccccccccccccccccc} \begin{aligned}&x = 0\end{aligned} &{}&\begin{aligned}&x + 4 = 0 \\ &x = - 4\end{aligned} \end{array}$$

$p:y = x + 4$

нула: $y = 0$

$x + 4 = 0$

$x = - 4$

$$P \cap p:\left\{ \begin{aligned}&y = {x^2} + 4x \\ &y = x + 4\end{aligned} \right.$$

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

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

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

$$\begin{array}{cccccccccccccccccccc} \begin{aligned}&{x_1} = - 4 \\ &{y_1} = 0 \\ &A\left( { - 4;0} \right)\end{aligned} &{}&\begin{aligned}&{x_2} = 1 \\ &{y_2} = 5 \\ &B\left( {1;5} \right)\end{aligned} \end{array}$$

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

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

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

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