Cho $y=f(x)$ là hàm số chẵn và liên tục trên $\mathbb{R}.$ Biết $\int\limits_{0}^{1}{f(x)\text{d}x=}\frac{1}{2}\int\limits_{1}^{2}{f(x)\text{d}x}=1.$ Giá trị của $\int\limits_{-2}^{2}{\frac{f(x)}{{{3}^{x}}+1}\text{d}x}$ bằng

Ta có: $\int\limits_{0}^{1}{f\left( x \right)dx=\frac{1}{2}\int\limits_{1}^{2}{f\left( x \right)dx=1\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx=1}}}$ và $\int\limits_{1}^{2}{f\left( x \right)dx=2.}$

$\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx+\int\limits_{1}^{2}{f\left( x \right)dx}=\int\limits_{0}^{2}{f\left( x \right)dx=3.}}$

Mặt khác: $\int\limits_{-2}^{2}{\frac{f\left( x \right)}{{{3}^{x}}+1}dx}=\int\limits_{-2}^{0}{\frac{f\left( x \right)}{{{3}^{x}}+1}dx+\int\limits_{0}^{2}{\frac{f\left( x \right)}{{{3}^{x}}+1}dx}}$ và $y=f\left( x \right)$ là hàm số chẵn, liên tục trên $R.$

$\Rightarrow f\left( -x \right)=f\left( x \right)\ \forall x\in R.$

Gọi $I=\int\limits_{-2}^{2}{\frac{f\left( x \right)}{{{3}^{x}}+1}\,\text{d}x}$, đặt $t=-\,x\Rightarrow \text{d}t=-\,\text{d}x$ và đổi cận $\left\{ \begin{align}  & x=-\,2\,\,\Rightarrow \,\,t=2 \\  & x=2\,\,\Rightarrow \,\,t=-\,2 \\ \end{align} \right..$

Suy ra $I=\int\limits_{2}^{-\,2}{\frac{f\left( -t \right)}{{{3}^{-t}}+1}\,\left( -\,\text{d}t \right)}=\int\limits_{-\,2}^{2}{\frac{f\left( t \right)}{\frac{1}{{{3}^{t}}}+1}\,\text{d}t}=\int\limits_{-\,2}^{2}{\frac{{{3}^{x}}f\left( x \right)}{{{3}^{x}}+1}\,\text{d}x}$

$\Rightarrow \,\,2I=\int\limits_{-\,2}^{2}{\frac{\left( {{3}^{x}}+1 \right)f\left( x \right)}{{{3}^{x}}+1}\,\text{d}x}=\int\limits_{-\,2}^{2}{f\left( x \right)\,\text{d}x}$

Do $f\left( x \right)$ là hàm chẵn nên suy ra $\int\limits_{-\,2}^{2}{f\left( x \right)\,\text{d}x}=2\int\limits_{0}^{2}{f\left( x \right)\,\text{d}x}$.

Vậy $I=\int\limits_{0}^{2}{f\left( x \right)\,\text{d}x}=\int\limits_{0}^{1}{f\left( x \right)\,\text{d}x}+\int\limits_{1}^{2}{f\left( x \right)\,\text{d}x}=3.$