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Câu hỏi: Cho hàm số chẵn $y=f\left( x \right)$ liên tục trên $\mathbb{R}$ và $\int\limits_{-1}^{1}{\dfrac{f\left( 2x \right)}{1+{{7}^{x}}}dx}=8$. Giá trị của $\int\limits_{0}^{2}{f\left( x \right)dx}$ bằng:
A. 8.
B. 2.
C. 1.
D. 16.
A. 8.
B. 2.
C. 1.
D. 16.
Ta có: $8=\int\limits_{-1}^{1}{\dfrac{f\left( 2x \right)}{1+{{7}^{x}}}dx=}\int\limits_{-1}^{0}{\dfrac{f\left( 2x \right)}{1+{{7}^{x}}}dx}+\int\limits_{0}^{1}{\dfrac{f\left( 2x \right)}{1+{{7}^{x}}}dx}\left( 1 \right)$
Xét $I=\int\limits_{-1}^{0}{\dfrac{f\left( 2x \right)}{1+{{7}^{x}}}dx}$ :
Đặt $t=-x\Rightarrow dt=-dx$. Đổi cận: $x=-1\Rightarrow t=1$ và $x=0\Rightarrow t=0$.
Khi đó $I=\int\limits_{1}^{0}{\dfrac{f\left( -2t \right)}{1+{{7}^{-t}}}\left( -dt \right)}=\int\limits_{0}^{1}{\dfrac{f\left( -2t \right)}{1+{{7}^{-t}}}dt}=\int\limits_{0}^{1}{\dfrac{{{7}^{t}}f\left( -2t \right)}{{{7}^{t}}+1}}dt$
Vì $y=f\left( x \right)$ là hàm chẵn trên $\mathbb{R}$ nên $f\left( -2t \right)=f\left( 2t \right),\forall t\in \mathbb{R}$.
Do đó $I=\int\limits_{0}^{1}{\dfrac{{{7}^{t}}f\left( 2t \right)}{{{7}^{t}}+1}dt}=\int\limits_{0}^{1}{\dfrac{{{7}^{x}}f\left( 2x \right)}{{{7}^{x}}+1}dx}$. Thay vào (1) thu được
$8=\int\limits_{0}^{1}{\dfrac{{{7}^{x}}f\left( 2x \right)}{{{7}^{x}}+1}dx+\int\limits_{0}^{1}{\dfrac{f\left( 2x \right)}{1+{{7}^{x}}}dx}=\int\limits_{0}^{1}{\dfrac{\left( {{7}^{x}}+1 \right)f\left( 2x \right)}{{{7}^{x}}+1}}}dx=\int\limits_{0}^{1}{f\left( 2x \right)dx}$.
$\Rightarrow \dfrac{1}{2}\int\limits_{0}^{1}{f\left( 2x \right)d\left( 2x \right)}=8\Rightarrow \int\limits_{0}^{2}{f\left( t \right)dt}=16$.
Xét $I=\int\limits_{-1}^{0}{\dfrac{f\left( 2x \right)}{1+{{7}^{x}}}dx}$ :
Đặt $t=-x\Rightarrow dt=-dx$. Đổi cận: $x=-1\Rightarrow t=1$ và $x=0\Rightarrow t=0$.
Khi đó $I=\int\limits_{1}^{0}{\dfrac{f\left( -2t \right)}{1+{{7}^{-t}}}\left( -dt \right)}=\int\limits_{0}^{1}{\dfrac{f\left( -2t \right)}{1+{{7}^{-t}}}dt}=\int\limits_{0}^{1}{\dfrac{{{7}^{t}}f\left( -2t \right)}{{{7}^{t}}+1}}dt$
Vì $y=f\left( x \right)$ là hàm chẵn trên $\mathbb{R}$ nên $f\left( -2t \right)=f\left( 2t \right),\forall t\in \mathbb{R}$.
Do đó $I=\int\limits_{0}^{1}{\dfrac{{{7}^{t}}f\left( 2t \right)}{{{7}^{t}}+1}dt}=\int\limits_{0}^{1}{\dfrac{{{7}^{x}}f\left( 2x \right)}{{{7}^{x}}+1}dx}$. Thay vào (1) thu được
$8=\int\limits_{0}^{1}{\dfrac{{{7}^{x}}f\left( 2x \right)}{{{7}^{x}}+1}dx+\int\limits_{0}^{1}{\dfrac{f\left( 2x \right)}{1+{{7}^{x}}}dx}=\int\limits_{0}^{1}{\dfrac{\left( {{7}^{x}}+1 \right)f\left( 2x \right)}{{{7}^{x}}+1}}}dx=\int\limits_{0}^{1}{f\left( 2x \right)dx}$.
$\Rightarrow \dfrac{1}{2}\int\limits_{0}^{1}{f\left( 2x \right)d\left( 2x \right)}=8\Rightarrow \int\limits_{0}^{2}{f\left( t \right)dt}=16$.
Đáp án D.