Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}7 - 4{x^2}\,\,\,khi\,\,\,0 \le x \le 1\\4 - {x^2}\,\,\,\,\,khi\,\,\,x > 1\end{array}

* Hướng dẫn giải

Xét các phương trình hoành độ giao điểm:

\(\begin{array}{l}
4 - {x^2} = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 2 \notin \left( {1; + \infty } \right)
\end{array} \right. \Leftrightarrow x = 2\\
7 - 4{x^2} = 0 \Leftrightarrow x =  \pm \frac{{\sqrt 7 }}{2} \notin \left[ {0;1} \right]\\
 \Rightarrow S = \int\limits_0^1 {\left| {7 - 4{x^3}} \right|dx + } \int\limits_1^2 {\left| {4 - {x^2}} \right|dx + \int\limits_2^3 {\left| {4 - {x^2}} \right|dx} } \\
 = \int\limits_0^1 {\left( {7 - 4{x^3}} \right)dx}  + \int\limits_1^2 {\left( {4 - {x^2}} \right)dx + } \int\limits_2^3 {\left( {{x^2} - 4} \right)dx} \\
 = \left( {7x - {x^4}} \right)\left| \begin{array}{l}
^1\\
_0
\end{array} \right. + \left( {4x - \frac{{{x^3}}}{3}} \right)\left| \begin{array}{l}
^2\\
_1
\end{array} \right. + \left( {\frac{{{x^3}}}{3} - 4x} \right)\left| \begin{array}{l}
^3\\
_2
\end{array} \right.\\
 = 7 - 1 + \frac{{16}}{3} - \frac{{11}}{3} - 3 + \frac{{16}}{3} = 10
\end{array}\)