Cho hàm số \(f(x) = \frac{a}{{{{\left( {x + 1} \right)}^3}}} + b.x.

* Hướng dẫn giải

\(\begin{array}{l}
\mathop \smallint \limits_0^1 f(x)dx = \mathop \smallint \limits_0^1 (\frac{a}{{{{\left( {x + 1} \right)}^3}}} + b.x.{e^x})dx\\
 = ( - \frac{a}{{2{{(x + 1)}^2}}} + b.x.{e^x} - b.{e^x})\left| \begin{array}{l}
1\\
0
\end{array} \right.\\
 = (\frac{{ - a}}{8} + b.e - b.e) - (\frac{{ - a}}{2} - b) = \frac{{3a}}{8} + b\\
 \Rightarrow \frac{{3a}}{8} + b = 5\begin{array}{*{20}{c}}
{}&{(1)}
\end{array}
\end{array}\)

\(\begin{array}{l}
f'\left( x \right) = \frac{{ - 3a}}{{{{(x + 1)}^2}}} + b.{e^x} + b.x.{e^x}\\
f'\left( 0 \right) =  - 22 \Leftrightarrow  - 3a + b =  - 22\begin{array}{*{20}{c}}
{}&{(2)}
\end{array}
\end{array}\)

Từ (1), (2) suy ra a = 8, b = 2 , S = a + b = 10.