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