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Câu hỏi: Cho hàm số $f(x)$ liên tục trên $\mathbb{R}$ thỏa mãn $f(x)={{e}^{x}}+\int\limits_{0}^{1}{tf(t)dt},\forall x\in \mathbb{R}$. Tính $f(\ln (5620))$.
A. 5622.
B. 5621.
C. 5620.
D. 5619.
A. 5622.
B. 5621.
C. 5620.
D. 5619.
Theo giả thiết, ta có: $f(x)={{e}^{x}}+c$, với $c=\int\limits_{0}^{1}{tf(t)dt}$ là hằng số. Khi đó:
$c=\int\limits_{0}^{1}{t\left( {{e}^{t}}+c \right)dt}=\int\limits_{0}^{1}{t{{e}^{t}}dt}+\int\limits_{0}^{1}{ctdt}={{I}_{1}}+{{I}_{2}}$, với ${{I}_{1}}=\int\limits_{0}^{1}{t{{e}^{t}}dt}$, ${{I}_{2}}=\int\limits_{0}^{1}{ctdt}$.
Vì ${{I}_{1}}=\int\limits_{0}^{1}{t{{e}^{t}}dt}=\int\limits_{0}^{1}{td({{e}^{t}})}=(t{{e}^{t}})\left| _{0}^{1} \right.-\int\limits_{0}^{1}{{{e}^{t}}dt}=e-({{e}^{t}})\left| _{0}^{1} \right.=e-(e-1)=1$, ${{I}_{2}}=\int\limits_{0}^{1}{ctdt}=(\dfrac{c{{t}^{2}}}{2})\left| _{0}^{1} \right.=\dfrac{c}{2}$ nên
$c={{I}_{1}}+{{I}_{2}}\Leftrightarrow c=1+\dfrac{c}{2}\Leftrightarrow c=2$. Vậy $f(x)={{e}^{x}}+2,\forall x\in \mathbb{R}$.
Do đó $f(\ln (5620))={{e}^{\ln (5620)}}+2=5620+2=5622$.
$c=\int\limits_{0}^{1}{t\left( {{e}^{t}}+c \right)dt}=\int\limits_{0}^{1}{t{{e}^{t}}dt}+\int\limits_{0}^{1}{ctdt}={{I}_{1}}+{{I}_{2}}$, với ${{I}_{1}}=\int\limits_{0}^{1}{t{{e}^{t}}dt}$, ${{I}_{2}}=\int\limits_{0}^{1}{ctdt}$.
Vì ${{I}_{1}}=\int\limits_{0}^{1}{t{{e}^{t}}dt}=\int\limits_{0}^{1}{td({{e}^{t}})}=(t{{e}^{t}})\left| _{0}^{1} \right.-\int\limits_{0}^{1}{{{e}^{t}}dt}=e-({{e}^{t}})\left| _{0}^{1} \right.=e-(e-1)=1$, ${{I}_{2}}=\int\limits_{0}^{1}{ctdt}=(\dfrac{c{{t}^{2}}}{2})\left| _{0}^{1} \right.=\dfrac{c}{2}$ nên
$c={{I}_{1}}+{{I}_{2}}\Leftrightarrow c=1+\dfrac{c}{2}\Leftrightarrow c=2$. Vậy $f(x)={{e}^{x}}+2,\forall x\in \mathbb{R}$.
Do đó $f(\ln (5620))={{e}^{\ln (5620)}}+2=5620+2=5622$.
Đáp án A.