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Câu hỏi: Xét hàm số $f\left( x \right)$ liên tục trên đoạn $\left[ 0;1 \right]$ và thỏa mãn $4x.f\left( {{x}^{2}} \right)+3f\left( 1-x \right)=\dfrac{x}{\sqrt{{{x}^{2}}+1}}$.
Tính $I=\int\limits_{0}^{1}{f\left( x \right)dx}$.
A. $I=\dfrac{2\sqrt{2}-1}{5}$.
B. $I=\dfrac{2\sqrt{2}-1}{10}$.
C. $I=\dfrac{\sqrt{2}-1}{5}$.
D. $I=\dfrac{\sqrt{2}-1}{10}$.
Tính $I=\int\limits_{0}^{1}{f\left( x \right)dx}$.
A. $I=\dfrac{2\sqrt{2}-1}{5}$.
B. $I=\dfrac{2\sqrt{2}-1}{10}$.
C. $I=\dfrac{\sqrt{2}-1}{5}$.
D. $I=\dfrac{\sqrt{2}-1}{10}$.
Ta có $\int\limits_{0}^{1}{4x.f\left( {{x}^{2}} \right)dx}+\int\limits_{0}^{1}{3f\left( 1-x \right)dx}=\int\limits_{0}^{1}{\dfrac{x}{\sqrt{{{x}^{2}}+1}}dx}$
$\int\limits_{0}^{1}{4x.f\left( {{x}^{2}} \right)dx}=2\int\limits_{0}^{1}{f\left( {{x}^{2}} \right)d\left( {{x}^{2}} \right)=2\int\limits_{0}^{1}{f\left( u \right)du}=2\int\limits_{0}^{1}{f\left( x \right)dx}}$.
$\int\limits_{0}^{1}{3f\left( 1-x \right)dx}=3\int\limits_{0}^{1}{f\left( v \right)f\left( 1-v \right)}=3\int\limits_{0}^{1}{f\left( v \right)dv}=3\int\limits_{0}^{1}{f\left( x \right)dx}$.
Do đó $5\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\dfrac{x}{\sqrt{{{x}^{2}}+1}}dx}=\dfrac{1}{2}\int\limits_{0}^{1}{\dfrac{1}{\sqrt{{{x}^{2}}+1}}d\left( {{x}^{2}}+1 \right)}=\left. \dfrac{1}{2}.2\sqrt{{{x}^{2}}+1} \right|_{0}^{1}=\sqrt{2-1}$
$\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{\sqrt{2}-1}{5}$.
$\int\limits_{0}^{1}{4x.f\left( {{x}^{2}} \right)dx}=2\int\limits_{0}^{1}{f\left( {{x}^{2}} \right)d\left( {{x}^{2}} \right)=2\int\limits_{0}^{1}{f\left( u \right)du}=2\int\limits_{0}^{1}{f\left( x \right)dx}}$.
$\int\limits_{0}^{1}{3f\left( 1-x \right)dx}=3\int\limits_{0}^{1}{f\left( v \right)f\left( 1-v \right)}=3\int\limits_{0}^{1}{f\left( v \right)dv}=3\int\limits_{0}^{1}{f\left( x \right)dx}$.
Do đó $5\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\dfrac{x}{\sqrt{{{x}^{2}}+1}}dx}=\dfrac{1}{2}\int\limits_{0}^{1}{\dfrac{1}{\sqrt{{{x}^{2}}+1}}d\left( {{x}^{2}}+1 \right)}=\left. \dfrac{1}{2}.2\sqrt{{{x}^{2}}+1} \right|_{0}^{1}=\sqrt{2-1}$
$\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\dfrac{\sqrt{2}-1}{5}$.
Đáp án C.