Cho \(f(x)\) là một hàm số liên tục trên \([-2;5]\) và \(\displaystyle\int\limits_{-2}^5f(x)\mathrm{\,d}x=8\), \(\displaystyle\int\limits_1^3f(x)\mathrm{\,d}x=-3\). Tính \(P=\displaystyle\int\limits_{-2}^1f(x)\mathrm{\,d}x+\displaystyle\int\limits_{3}^5f(x)\mathrm{\,d}x\).
![]() | \(P=5\) |
![]() | \(P=-11\) |
![]() | \(P=11\) |
![]() | \(P=-5\) |
Chọn phương án C.
$$\begin{eqnarray*}
&\displaystyle\int\limits_{-2}^5 f(x)\mathrm{\,d}x &=\int\limits_{-2}^1 f(x)\mathrm{\,d}x+\int\limits_{1}^3 f(x)\mathrm{\,d}x+\int\limits_{3}^5 f(x)\mathrm{\,d}x\\
\Leftrightarrow &8 &=\int\limits_{-2}^1 f(x)\mathrm{\,d}x-3+\int\limits_{3}^5 f(x)\mathrm{\,d}x\\
\Leftrightarrow &11 &=\int\limits_{-2}^1 f(x)\mathrm{\,d}x+\int\limits_{3}^5 f(x)\mathrm{\,d}x\\
\Leftrightarrow &11 &=P.
\end{eqnarray*}$$