Google
Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên ${ \mathbb{R} }$ thỏa mãn $f\left( x \right)={{e}^{x}}+\int\limits_{0}^{1}{t}f\left( t \right)dt,\forall x\in \mathbb{R}$. Tính $f\left( \ln \left( 5620 \right) \right)$
A. 5622.
B. 5621.
C. 5620.
D. 5619.
A. 5622.
B. 5621.
C. 5620.
D. 5619.
Đặt $I=$ $\int\limits_{0}^{1}{t}f\left( t \right)dt$ (với $I$ là số thực). Khi đó $f\left( x \right)={{e}^{x}}+I\Rightarrow f\left( t \right)={{e}^{t}}+I$
Ta có: $I=\int\limits_{0}^{1}{t}f\left( t \right)dt=\int\limits_{0}^{1}{t}\left( {{e}^{t}}+I \right)dt=\int\limits_{0}^{1}{{{e}^{t}}.t.}dt+I\int\limits_{0}^{1}{t}dt=\int\limits_{0}^{1}{t.}d\left( {{e}^{t}} \right)+I\int\limits_{0}^{1}{t}dt=t.{{e}^{t}}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{{{e}^{t}}}dt+I\dfrac{{{t}^{2}}}{2}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.$
$I=e-{{e}^{t}}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.+\dfrac{I}{2}=1+\dfrac{I}{2}\Rightarrow \dfrac{I}{2}=1\Leftrightarrow I=2\Rightarrow f\left( x \right)={{e}^{x}}+2$.
Khi đó: $f\left( \ln \left( 5620 \right) \right)={{e}^{\ln 5620}}+2=5622$
Ta có: $I=\int\limits_{0}^{1}{t}f\left( t \right)dt=\int\limits_{0}^{1}{t}\left( {{e}^{t}}+I \right)dt=\int\limits_{0}^{1}{{{e}^{t}}.t.}dt+I\int\limits_{0}^{1}{t}dt=\int\limits_{0}^{1}{t.}d\left( {{e}^{t}} \right)+I\int\limits_{0}^{1}{t}dt=t.{{e}^{t}}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{{{e}^{t}}}dt+I\dfrac{{{t}^{2}}}{2}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.$
$I=e-{{e}^{t}}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.+\dfrac{I}{2}=1+\dfrac{I}{2}\Rightarrow \dfrac{I}{2}=1\Leftrightarrow I=2\Rightarrow f\left( x \right)={{e}^{x}}+2$.
Khi đó: $f\left( \ln \left( 5620 \right) \right)={{e}^{\ln 5620}}+2=5622$
Đáp án A.