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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 1;4 \right]$, đồng biến trên đoạn $\left[ 1;4 \right]$ và thỏa mãn đẳng thức $x+2x.f\left( x \right)$ $={{\left[ {f}'\left( x \right) \right]}^{2}}$, $\forall x\in \left[ 1;4 \right]$.
Biết rằng $f\left( 1 \right)=\dfrac{3}{2}$, tính $I=\int\limits_{1}^{4}{f\left( x \right)\text{d}x}$ ?
A. $I=\dfrac{1186}{45}$.
B. $I=\dfrac{1174}{45}$.
C. $I=\dfrac{1222}{45}$.
D. $I=\dfrac{1201}{45}$.
Biết rằng $f\left( 1 \right)=\dfrac{3}{2}$, tính $I=\int\limits_{1}^{4}{f\left( x \right)\text{d}x}$ ?
A. $I=\dfrac{1186}{45}$.
B. $I=\dfrac{1174}{45}$.
C. $I=\dfrac{1222}{45}$.
D. $I=\dfrac{1201}{45}$.
Ta có $x+2x.f\left( x \right)$ $={{\left[ {f}'\left( x \right) \right]}^{2}}$ $\Rightarrow \sqrt{x}.\sqrt{1+2f\left( x \right)}={f}'\left( x \right)$ $\Rightarrow \dfrac{{f}'\left( x \right)}{\sqrt{1+2f\left( x \right)}}=\sqrt{x}$, $\forall x\in \left[ 1;4 \right]$.
Suy ra $\int{\dfrac{{f}'\left( x \right)}{\sqrt{1+2f\left( x \right)}}\text{d}x}=\int{\sqrt{x}\text{d}x}+C$ $\Leftrightarrow \int{\dfrac{\text{d}f\left( x \right)}{\sqrt{1+2f\left( x \right)}}\text{d}x}=\int{\sqrt{x}\text{d}x}+C$
$\Rightarrow \sqrt{1+2f\left( x \right)}=\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}+C$. Mà $f\left( 1 \right)=\dfrac{3}{2}$ $\Rightarrow C=\dfrac{4}{3}$. Vậy $f\left( x \right)=\dfrac{{{\left( \dfrac{2}{3}{{x}^{\dfrac{3}{2}}}+\dfrac{4}{3} \right)}^{2}}-1}{2}$.
Vậy $I=\int\limits_{1}^{4}{f\left( x \right)\text{d}x}=\dfrac{1186}{45}$.
Suy ra $\int{\dfrac{{f}'\left( x \right)}{\sqrt{1+2f\left( x \right)}}\text{d}x}=\int{\sqrt{x}\text{d}x}+C$ $\Leftrightarrow \int{\dfrac{\text{d}f\left( x \right)}{\sqrt{1+2f\left( x \right)}}\text{d}x}=\int{\sqrt{x}\text{d}x}+C$
$\Rightarrow \sqrt{1+2f\left( x \right)}=\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}+C$. Mà $f\left( 1 \right)=\dfrac{3}{2}$ $\Rightarrow C=\dfrac{4}{3}$. Vậy $f\left( x \right)=\dfrac{{{\left( \dfrac{2}{3}{{x}^{\dfrac{3}{2}}}+\dfrac{4}{3} \right)}^{2}}-1}{2}$.
Vậy $I=\int\limits_{1}^{4}{f\left( x \right)\text{d}x}=\dfrac{1186}{45}$.
Đáp án A.