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Câu hỏi: Xét hàm số $f(x)$ liên tục trên $\left[ 0;1 \right]$ và thỏa mãn điều kiện $4x.f\left( {{x}^{2}} \right)+3f\left( 1-x \right)=\sqrt{1-{{x}^{2}}}$. Tích phân $I=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$ bằng:
A. $I=\dfrac{\pi }{16}$
B. $I=\dfrac{\pi }{4}$
C. $I=\dfrac{\pi }{6}$
D. $I=\dfrac{\pi }{20}$
A. $I=\dfrac{\pi }{16}$
B. $I=\dfrac{\pi }{4}$
C. $I=\dfrac{\pi }{6}$
D. $I=\dfrac{\pi }{20}$
Xét $4x.f\left( {{x}^{2}} \right)+3f(1-x)=\sqrt{1-{{x}^{2}}}$
Suy ra: $\underbrace{\int\limits_{0}^{1}{\left[ 4x.f\left( {{x}^{2}} \right) \right]} \text{d}x}_{{{I}_{1}}}+\underbrace{\int\limits_{0}^{1}{3f\left( 1-x \right)} \text{d}x}_{{{I}_{2}}}=\int\limits_{0}^{1}{\sqrt{1-{{x}^{2}}}}\text{d}x \left( * \right)$.
Xét ${{I}_{1}}=\int\limits_{0}^{1}{\left[ 4x.f\left( {{x}^{2}} \right) \right]} \text{d}x$
Đặt $t={{x}^{2}}\Rightarrow dt=2xdx$.
Đôi cận $x=0\Rightarrow t=0$ ; $x=1\Rightarrow t=1$.
Suy ra: ${{I}_{1}}=2.\int\limits_{0}^{1}{f\left( t \right)}dt=2.\int\limits_{0}^{1}{f\left( x \right)}dx$.
Xét ${{I}_{2}}=\int\limits_{0}^{1}{3f(1-x)} \text{d}x $.
Đặt $t=1-x\Rightarrow dt=-dx$.
Đôi cận $x=0\Rightarrow t=1$ ; $x=1\Rightarrow t=0$.
Suy ra: ${{I}_{1}}=3.\int\limits_{1}^{0}{f\left( t \right)}\left( -dt \right)=3.\int\limits_{0}^{1}{f\left( t \right)}dt=3.\int\limits_{0}^{1}{f\left( x \right)}dx$.
Thay vào $\left( * \right)$ ta được:
$2.\int\limits_{0}^{1}{f\left( x \right)}dx+3.\int\limits_{0}^{1}{f\left( x \right)}dx=\dfrac{\pi }{4} \Leftrightarrow 5.\int\limits_{0}^{1}{f\left( x \right)}dx=\dfrac{\pi }{4}\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)}dx=\dfrac{\pi }{20}$.
Suy ra: $\underbrace{\int\limits_{0}^{1}{\left[ 4x.f\left( {{x}^{2}} \right) \right]} \text{d}x}_{{{I}_{1}}}+\underbrace{\int\limits_{0}^{1}{3f\left( 1-x \right)} \text{d}x}_{{{I}_{2}}}=\int\limits_{0}^{1}{\sqrt{1-{{x}^{2}}}}\text{d}x \left( * \right)$.
Xét ${{I}_{1}}=\int\limits_{0}^{1}{\left[ 4x.f\left( {{x}^{2}} \right) \right]} \text{d}x$
Đặt $t={{x}^{2}}\Rightarrow dt=2xdx$.
Đôi cận $x=0\Rightarrow t=0$ ; $x=1\Rightarrow t=1$.
Suy ra: ${{I}_{1}}=2.\int\limits_{0}^{1}{f\left( t \right)}dt=2.\int\limits_{0}^{1}{f\left( x \right)}dx$.
Xét ${{I}_{2}}=\int\limits_{0}^{1}{3f(1-x)} \text{d}x $.
Đặt $t=1-x\Rightarrow dt=-dx$.
Đôi cận $x=0\Rightarrow t=1$ ; $x=1\Rightarrow t=0$.
Suy ra: ${{I}_{1}}=3.\int\limits_{1}^{0}{f\left( t \right)}\left( -dt \right)=3.\int\limits_{0}^{1}{f\left( t \right)}dt=3.\int\limits_{0}^{1}{f\left( x \right)}dx$.
Thay vào $\left( * \right)$ ta được:
$2.\int\limits_{0}^{1}{f\left( x \right)}dx+3.\int\limits_{0}^{1}{f\left( x \right)}dx=\dfrac{\pi }{4} \Leftrightarrow 5.\int\limits_{0}^{1}{f\left( x \right)}dx=\dfrac{\pi }{4}\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)}dx=\dfrac{\pi }{20}$.
Đáp án D.