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Пр.4)   Израчунај површину између криве $y = 2{x^2}$, праве $x = 0$

            и тангенте дате криве у тачки $A\left( {2,{y_0}} \right)$.

$y = 2{x^2}$

$x{\text{ }} = {\text{ }}0$

$A\left( {2,{y_0}} \right)$

$A \in P:{y_0} = 2 \cdot {2^2}$

${y_0} = 8$

$A\left( {2;8} \right)$

$t:y - {y_0} = k\left( {x - {x_0}} \right)$

$k = f'\left( {{x_0}} \right)$

$y' = 2 \cdot 2x$

$y' = 4x$

$y'\left( 2 \right) = 4 \cdot 2$

$y'\left( 2 \right) = 8$

$t:y - 8 = 8\left( {x - 2} \right)$

$t:y = 8x - 16 + 8$

$t:y = 8x - 8$

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

$y = 0;x = 0$

$x = 0$

$t:y = 8x - 8$

\[\begin{array}{*{20}{c}}
{y = 0}&{}&\begin{gathered}
8x - 8 = 0 \hfill \\
x = 1 \hfill \\
\end{gathered}
\end{array}\]

$P = \int\limits_0^2 {2{x^2}dx - \int\limits_1^2 {\left( {8x - 8} \right)dx - \int\limits_0^1 {\left( {8x - 8} \right)dx = } } } \int\limits_0^2 {2{x^2}dx}  - \left( {\int\limits_0^2 {\left( {8x - 8} \right)dx} } \right) = $

$ = \int\limits_0^2 {\left( {2{x^2} - 8x + 8} \right)dx}  = \left. {\left( {2\frac{{{x^3}}}{3} - 8\frac{{{x^2}}}{2} + 8x} \right)} \right|_0^2 = \frac{2}{3} \cdot 8 - 4 \cdot {2^2} + 8 \cdot 2 - 0 = $

$ = \frac{{16}}{3} - 16 + 16 = \frac{{16}}{3}$