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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên đoạn $\left[ 0;1 \right]$ và thỏa mãn $\sqrt{{{x}^{3}}+1}\left[ 4x{f}'\left( 1-x \right)-f\left( x \right) \right]={{x}^{5}}$. Tích phân $I=\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$ có kết quả dạng $\dfrac{a-b\sqrt{2}}{c}$, ( $a$, $b$, $c$ $\in {{\mathbb{Z}}^{+}}$, $\dfrac{a}{c}$, $\dfrac{b}{c}$ là phân số tối giản). Giá trị $T=a-2b+3c$ bằng:
A. $89$.
B. $27$.
C. $35$.
D. $81$.
A. $89$.
B. $27$.
C. $35$.
D. $81$.
Thay $x=0$ vào $\sqrt{{{x}^{3}}+1}\left[ 4x{f}'\left( 1-x \right)-f\left( x \right) \right]={{x}^{5}}$, ta có $f\left( 0 \right)=0$.
$\sqrt{{{x}^{3}}+1}\left[ 4x{f}'\left( 1-x \right)-f\left( x \right) \right]={{x}^{5}}\Leftrightarrow 4x{f}'\left( 1-x \right)-f\left( x \right)=\dfrac{{{x}^{5}}}{\sqrt{{{x}^{3}}+1}}$
$\Rightarrow \int\limits_{0}^{1}{4x{f}'\left( 1-x \right)\text{d}x}-\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\dfrac{{{x}^{5}}}{\sqrt{{{x}^{3}}+1}}\text{d}x}$
Xét tích phân $\int\limits_{0}^{1}{4x{f}'\left( 1-x \right)\text{d}x}$
Ta có $\int\limits_{0}^{1}{4x{f}'\left( 1-x \right)\text{d}x}=\int\limits_{0}^{1}{4x\text{d}\left[ -f\left( 1-x \right) \right]}=\left. -4xf\left( 1-x \right) \right|_{0}^{1}+4\int\limits_{0}^{1}{f\left( 1-x \right)\text{d}x}$
$\Leftrightarrow \int\limits_{0}^{1}{4x{f}'\left( 1-x \right)\text{d}x}=-4f\left( 0 \right)+4\int\limits_{0}^{1}{f\left( x \right)\text{d}x}\Leftrightarrow \int\limits_{0}^{1}{4x{f}'\left( 1-x \right)\text{d}x}=4\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$
Xét tích phân $\int\limits_{0}^{1}{\dfrac{{{x}^{5}}}{\sqrt{{{x}^{3}}+1}}\text{d}x}$
Đặt $t=\sqrt{{{x}^{3}}+1}\Leftrightarrow {{t}^{2}}={{x}^{3}}+1\Rightarrow \left\{ \begin{matrix}
2t\text{d}t=3{{x}^{2}}\text{d}x \\
x=0\Rightarrow t=1 \\
x=1\Rightarrow t=\sqrt{2} \\
\end{matrix} \right.$, khi đó:
$\int\limits_{0}^{1}{\dfrac{{{x}^{5}}}{\sqrt{{{x}^{3}}+1}}\text{d}x}=\dfrac{2}{3}\int\limits_{1}^{\sqrt{2}}{\left( {{t}^{2}}-1 \right)\text{d}t}=\left. \dfrac{2}{3}\left( \dfrac{{{t}^{3}}}{3}-t \right) \right|_{1}^{\sqrt{2}}=\dfrac{4-2\sqrt{2}}{9}$.
Khi đó $\int\limits_{0}^{1}{4x{f}'\left( 1-x \right)\text{d}x}-\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\dfrac{{{x}^{5}}}{\sqrt{{{x}^{3}}+1}}\text{d}x}$
$\Leftrightarrow 3\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{4-2\sqrt{2}}{9}\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{4-2\sqrt{2}}{27}$ $\Rightarrow \left\{ \begin{matrix}
a=4 \\
b=2 \\
c=27 \\
\end{matrix} \right.\Rightarrow T=a-2b+3c=81$.
$\sqrt{{{x}^{3}}+1}\left[ 4x{f}'\left( 1-x \right)-f\left( x \right) \right]={{x}^{5}}\Leftrightarrow 4x{f}'\left( 1-x \right)-f\left( x \right)=\dfrac{{{x}^{5}}}{\sqrt{{{x}^{3}}+1}}$
$\Rightarrow \int\limits_{0}^{1}{4x{f}'\left( 1-x \right)\text{d}x}-\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\dfrac{{{x}^{5}}}{\sqrt{{{x}^{3}}+1}}\text{d}x}$
Xét tích phân $\int\limits_{0}^{1}{4x{f}'\left( 1-x \right)\text{d}x}$
Ta có $\int\limits_{0}^{1}{4x{f}'\left( 1-x \right)\text{d}x}=\int\limits_{0}^{1}{4x\text{d}\left[ -f\left( 1-x \right) \right]}=\left. -4xf\left( 1-x \right) \right|_{0}^{1}+4\int\limits_{0}^{1}{f\left( 1-x \right)\text{d}x}$
$\Leftrightarrow \int\limits_{0}^{1}{4x{f}'\left( 1-x \right)\text{d}x}=-4f\left( 0 \right)+4\int\limits_{0}^{1}{f\left( x \right)\text{d}x}\Leftrightarrow \int\limits_{0}^{1}{4x{f}'\left( 1-x \right)\text{d}x}=4\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$
Xét tích phân $\int\limits_{0}^{1}{\dfrac{{{x}^{5}}}{\sqrt{{{x}^{3}}+1}}\text{d}x}$
Đặt $t=\sqrt{{{x}^{3}}+1}\Leftrightarrow {{t}^{2}}={{x}^{3}}+1\Rightarrow \left\{ \begin{matrix}
2t\text{d}t=3{{x}^{2}}\text{d}x \\
x=0\Rightarrow t=1 \\
x=1\Rightarrow t=\sqrt{2} \\
\end{matrix} \right.$, khi đó:
$\int\limits_{0}^{1}{\dfrac{{{x}^{5}}}{\sqrt{{{x}^{3}}+1}}\text{d}x}=\dfrac{2}{3}\int\limits_{1}^{\sqrt{2}}{\left( {{t}^{2}}-1 \right)\text{d}t}=\left. \dfrac{2}{3}\left( \dfrac{{{t}^{3}}}{3}-t \right) \right|_{1}^{\sqrt{2}}=\dfrac{4-2\sqrt{2}}{9}$.
Khi đó $\int\limits_{0}^{1}{4x{f}'\left( 1-x \right)\text{d}x}-\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\dfrac{{{x}^{5}}}{\sqrt{{{x}^{3}}+1}}\text{d}x}$
$\Leftrightarrow 3\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{4-2\sqrt{2}}{9}\Leftrightarrow \int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{4-2\sqrt{2}}{27}$ $\Rightarrow \left\{ \begin{matrix}
a=4 \\
b=2 \\
c=27 \\
\end{matrix} \right.\Rightarrow T=a-2b+3c=81$.
Đáp án D.
