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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 1;8 \right]$ và thỏa mãn
A. $\dfrac{257\ln 2}{2}$.
B. $\dfrac{257\ln 2}{4}$ .
C. $160$ .
D. $\dfrac{639}{4}$.
$\int\limits_{1}^{2}{{{\left[ f\left( {{x}^{3}} \right) \right]}^{2}}\text{d}x+2\int\limits_{1}^{2}{f\left( {{x}^{3}} \right)\text{d}x-\dfrac{4}{3}\int\limits_{1}^{8}{f\left( x \right)\text{d}x=-\dfrac{247}{15}}}}$.
Giả sử rằng $F\left( x \right)$ là một nguyên hàm của hàm số $f\left( x \right)$ trên $\left[ 1;8 \right]$. Tích phân $\int\limits_{1}^{8}{x\cdot {F}'\left( x \right)\text{d}x}$ bằngA. $\dfrac{257\ln 2}{2}$.
B. $\dfrac{257\ln 2}{4}$ .
C. $160$ .
D. $\dfrac{639}{4}$.
Xét $I=\int\limits_{1}^{2}{{{\left[ f\left( {{x}^{3}} \right) \right]}^{2}}\text{d}x+2\int\limits_{1}^{2}{f\left( {{x}^{3}} \right)\text{d}x}}=\int\limits_{1}^{2}{{{\left[ f\left( {{x}^{3}} \right)+1 \right]}^{2}}\text{d}x}-\int\limits_{1}^{2}{\text{d}x}=\int\limits_{1}^{2}{{{\left[ f\left( {{x}^{3}} \right)+1 \right]}^{2}}\text{d}x}-1$.
Đặt $t={{x}^{3}}\Rightarrow \text{d}t=3{{x}^{2}}\text{d}x\Leftrightarrow \text{d}x=\dfrac{\text{d}t}{3\sqrt[3]{{{t}^{2}}}}$.
Với $x=1\Rightarrow t=1$ ;
$x=2\Rightarrow t=8$.
Ta có $I=\dfrac{1}{3}\int\limits_{1}^{8}{\dfrac{{{\left[ f\left( t \right)+1 \right]}^{2}}}{\sqrt[3]{{{t}^{2}}}}\text{d}t}-1=\dfrac{1}{3}\int\limits_{1}^{8}{{{\left[ \dfrac{f\left( x \right)+1}{\sqrt[3]{x}} \right]}^{2}}\text{d}x}-1$.
Do đó
$\int\limits_{1}^{2}{{{\left[ f\left( {{x}^{3}} \right) \right]}^{2}}\text{d}x+2\int\limits_{1}^{2}{f\left( {{x}^{3}} \right)\text{d}x-\dfrac{4}{3}\int\limits_{1}^{8}{f\left( x \right)\text{d}x=-\dfrac{247}{15}}}}$
$\Leftrightarrow \dfrac{1}{3}\int\limits_{1}^{8}{{{\left[ \dfrac{f\left( x \right)+1}{\sqrt[3]{x}} \right]}^{2}}\text{d}x}-1-\dfrac{4}{3}\int\limits_{1}^{8}{f\left( x \right)\text{d}x=\dfrac{-247}{15}}$
$\Leftrightarrow \int\limits_{1}^{8}{{{\left[ \dfrac{f\left( x \right)+1}{\sqrt[3]{x}} \right]}^{2}}\text{d}x}-3-4\int\limits_{1}^{8}{f\left( x \right)\text{d}x=\dfrac{-247}{5}}$
$\Leftrightarrow \int\limits_{1}^{8}{\left[ {{\left( \dfrac{f\left( x \right)+1}{\sqrt[3]{x}} \right)}^{2}}-4f\left( x \right) \right]\text{d}x}=\dfrac{-232}{5}$
$\Leftrightarrow \int\limits_{1}^{8}{\left[ {{\left( \dfrac{f\left( x \right)+1}{\sqrt[3]{x}} \right)}^{2}}-2\cdot \dfrac{f\left( x \right)+1}{\sqrt[3]{x}}\cdot 2\sqrt[3]{x}+4\sqrt[3]{{{x}^{2}}} \right]\text{d}x+\int\limits_{1}^{8}{\left[ -4\sqrt[3]{{{x}^{2}}}+4 \right]\text{d}x}}=\dfrac{-232}{5}$
$\Leftrightarrow \int\limits_{1}^{8}{{{\left[ \dfrac{f\left( x \right)+1}{\sqrt[3]{x}}-2\sqrt[3]{x} \right]}^{2}}\text{d}x=0}$, do $\int\limits_{1}^{8}{\left[ -4\sqrt[3]{{{x}^{2}}}+4 \right]\text{d}x}=\dfrac{-232}{5}$
$\Rightarrow \dfrac{f\left( x \right)+1}{\sqrt[3]{x}}-2\sqrt[3]{x}=0\Rightarrow f\left( x \right)=2\sqrt[3]{{{x}^{2}}}-1={F}'\left( x \right)$.
Suy ra $\int\limits_{1}^{8}{x\cdot {F}'\left( x \right)\text{d}x=\int\limits_{1}^{8}{x\left[ 2\sqrt[3]{{{x}^{2}}}-1 \right]\text{d}x}}$ $=2\int\limits_{1}^{8}{{{x}^{\dfrac{5}{3}}}\text{d}x-\int\limits_{1}^{8}{x\text{d}x=\dfrac{639}{4}}}$.
Đặt $t={{x}^{3}}\Rightarrow \text{d}t=3{{x}^{2}}\text{d}x\Leftrightarrow \text{d}x=\dfrac{\text{d}t}{3\sqrt[3]{{{t}^{2}}}}$.
Với $x=1\Rightarrow t=1$ ;
$x=2\Rightarrow t=8$.
Ta có $I=\dfrac{1}{3}\int\limits_{1}^{8}{\dfrac{{{\left[ f\left( t \right)+1 \right]}^{2}}}{\sqrt[3]{{{t}^{2}}}}\text{d}t}-1=\dfrac{1}{3}\int\limits_{1}^{8}{{{\left[ \dfrac{f\left( x \right)+1}{\sqrt[3]{x}} \right]}^{2}}\text{d}x}-1$.
Do đó
$\int\limits_{1}^{2}{{{\left[ f\left( {{x}^{3}} \right) \right]}^{2}}\text{d}x+2\int\limits_{1}^{2}{f\left( {{x}^{3}} \right)\text{d}x-\dfrac{4}{3}\int\limits_{1}^{8}{f\left( x \right)\text{d}x=-\dfrac{247}{15}}}}$
$\Leftrightarrow \dfrac{1}{3}\int\limits_{1}^{8}{{{\left[ \dfrac{f\left( x \right)+1}{\sqrt[3]{x}} \right]}^{2}}\text{d}x}-1-\dfrac{4}{3}\int\limits_{1}^{8}{f\left( x \right)\text{d}x=\dfrac{-247}{15}}$
$\Leftrightarrow \int\limits_{1}^{8}{{{\left[ \dfrac{f\left( x \right)+1}{\sqrt[3]{x}} \right]}^{2}}\text{d}x}-3-4\int\limits_{1}^{8}{f\left( x \right)\text{d}x=\dfrac{-247}{5}}$
$\Leftrightarrow \int\limits_{1}^{8}{\left[ {{\left( \dfrac{f\left( x \right)+1}{\sqrt[3]{x}} \right)}^{2}}-4f\left( x \right) \right]\text{d}x}=\dfrac{-232}{5}$
$\Leftrightarrow \int\limits_{1}^{8}{\left[ {{\left( \dfrac{f\left( x \right)+1}{\sqrt[3]{x}} \right)}^{2}}-2\cdot \dfrac{f\left( x \right)+1}{\sqrt[3]{x}}\cdot 2\sqrt[3]{x}+4\sqrt[3]{{{x}^{2}}} \right]\text{d}x+\int\limits_{1}^{8}{\left[ -4\sqrt[3]{{{x}^{2}}}+4 \right]\text{d}x}}=\dfrac{-232}{5}$
$\Leftrightarrow \int\limits_{1}^{8}{{{\left[ \dfrac{f\left( x \right)+1}{\sqrt[3]{x}}-2\sqrt[3]{x} \right]}^{2}}\text{d}x=0}$, do $\int\limits_{1}^{8}{\left[ -4\sqrt[3]{{{x}^{2}}}+4 \right]\text{d}x}=\dfrac{-232}{5}$
$\Rightarrow \dfrac{f\left( x \right)+1}{\sqrt[3]{x}}-2\sqrt[3]{x}=0\Rightarrow f\left( x \right)=2\sqrt[3]{{{x}^{2}}}-1={F}'\left( x \right)$.
Suy ra $\int\limits_{1}^{8}{x\cdot {F}'\left( x \right)\text{d}x=\int\limits_{1}^{8}{x\left[ 2\sqrt[3]{{{x}^{2}}}-1 \right]\text{d}x}}$ $=2\int\limits_{1}^{8}{{{x}^{\dfrac{5}{3}}}\text{d}x-\int\limits_{1}^{8}{x\text{d}x=\dfrac{639}{4}}}$.
Đáp án D.