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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm và đồng biến trên $\left[ 1;4 \right]$, thoả mãn $x+2x.f\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}},\forall x\in \left[ 1;4 \right]$. Biết rằng $f\left( 1 \right)=\dfrac{3}{2}$. Tính tích phân $I=\int\limits_{1}^{4}{f\left( x \right)}dx$ ?
A. $\dfrac{9}{2}$.
B. $\dfrac{1187}{45}$.
C. $\dfrac{1188}{45}$
D. $\dfrac{1186}{45}$.
A. $\dfrac{9}{2}$.
B. $\dfrac{1187}{45}$.
C. $\dfrac{1188}{45}$
D. $\dfrac{1186}{45}$.
Vì $f\left( x \right)$ có đạo hàm và đồng biến trên $\left[ 1;4 \right]$ suy ra ${f}'\left( x \right)>0,\forall x\in \left[ 1;4 \right]$. Khi đó
$x+2x.f\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}}\Leftrightarrow x\left[ 1+2f\left( x \right) \right]={{\left[ {f}'\left( x \right) \right]}^{2}}\Leftrightarrow \sqrt{x}=\dfrac{{f}'\left( x \right)}{\sqrt{1+2f\left( x \right)}}={{\left( \sqrt{1+2f\left( x \right)} \right)}^{\prime }}$
$\Rightarrow \sqrt{1+2f\left( x \right)}=\dfrac{2}{3}\sqrt{{{x}^{3}}}+C$.
Mà $f\left( 1 \right)=\dfrac{3}{2}\Rightarrow \sqrt{1+3}=\dfrac{2}{3}+C\Rightarrow C=\dfrac{4}{3}\Rightarrow f\left( x \right)=\dfrac{{{\left( \dfrac{2}{3}\sqrt{{{x}^{3}}}+\dfrac{4}{3} \right)}^{2}}-1}{2}$
$\Rightarrow I=\int\limits_{1}^{4}{f\left( x \right)}dx=\int\limits_{1}^{4}{\dfrac{{{\left( \dfrac{2}{3}\sqrt{{{x}^{3}}}+\dfrac{4}{3} \right)}^{2}}-1}{2}dx}=\dfrac{1186}{45}$.
$x+2x.f\left( x \right)={{\left[ {f}'\left( x \right) \right]}^{2}}\Leftrightarrow x\left[ 1+2f\left( x \right) \right]={{\left[ {f}'\left( x \right) \right]}^{2}}\Leftrightarrow \sqrt{x}=\dfrac{{f}'\left( x \right)}{\sqrt{1+2f\left( x \right)}}={{\left( \sqrt{1+2f\left( x \right)} \right)}^{\prime }}$
$\Rightarrow \sqrt{1+2f\left( x \right)}=\dfrac{2}{3}\sqrt{{{x}^{3}}}+C$.
Mà $f\left( 1 \right)=\dfrac{3}{2}\Rightarrow \sqrt{1+3}=\dfrac{2}{3}+C\Rightarrow C=\dfrac{4}{3}\Rightarrow f\left( x \right)=\dfrac{{{\left( \dfrac{2}{3}\sqrt{{{x}^{3}}}+\dfrac{4}{3} \right)}^{2}}-1}{2}$
$\Rightarrow I=\int\limits_{1}^{4}{f\left( x \right)}dx=\int\limits_{1}^{4}{\dfrac{{{\left( \dfrac{2}{3}\sqrt{{{x}^{3}}}+\dfrac{4}{3} \right)}^{2}}-1}{2}dx}=\dfrac{1186}{45}$.
Đáp án D.