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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 0;1 \right]$, $f\left( x \right)$ và ${f}'\left( x \right)$ đều nhận giá trị dương trên đoạn $\left[ 0;1 \right]$. Biết rằng $\int\limits_{0}^{1}{\left[ {f}'\left( x \right).{{\left[ f\left( x \right) \right]}^{2}}+4 \right]dx}=4\int\limits_{0}^{1}{\sqrt{{f}'\left( x \right)}.f\left( x \right)dx}$ và $f\left( 0 \right)=3$. Tích phân
$\int\limits_{0}^{1}{{{\left[ f\left( x \right) \right]}^{3}}dx}$ bằng
A. 30.
B. 10.
C. 21.
D. 19.
$\int\limits_{0}^{1}{{{\left[ f\left( x \right) \right]}^{3}}dx}$ bằng
A. 30.
B. 10.
C. 21.
D. 19.
Ta có $\int\limits_{0}^{1}{\left[ {f}'\left( x \right).{{\left[ f\left( x \right) \right]}^{2}}+4 \right]dx}-4\int\limits_{0}^{1}{\sqrt{{f}'\left( x \right)}.f\left( x \right)dx}=0$
$\Rightarrow \int\limits_{0}^{1}{\left[ {f}'\left( x \right).{{\left[ f\left( x \right) \right]}^{2}}-4\sqrt{{f}'\left( x \right)}.f\left( x \right)+4 \right]dx}=0\Rightarrow \int\limits_{0}^{1}{{{\left[ \sqrt{{f}'\left( x \right)}.f\left( x \right)-2 \right]}^{2}}dx}=0$
$\Rightarrow \sqrt{{f}'\left( x \right)}.f\left( x \right)-2=0\Rightarrow {{\left[ f\left( x \right) \right]}^{2}}.{f}'\left( x \right)=2\Rightarrow \int\limits_{{}}^{{}}{{{\left[ f\left( x \right) \right]}^{2}}.{f}'\left( x \right)dx}=2x+{{C}_{1}}$
$\Rightarrow \int\limits_{{}}^{{}}{{{\left[ f\left( x \right) \right]}^{2}}d\left[ f\left( x \right) \right]}=2x+{{C}_{1}}\Rightarrow \dfrac{{{\left[ f\left( x \right) \right]}^{3}}}{3}=2x+{{C}_{2}}$
Mà $f\left( 0 \right)=3\Rightarrow {{C}_{2}}=9\Rightarrow {{\left[ f\left( x \right) \right]}^{3}}=3\left( 2x+9 \right)\Rightarrow \int\limits_{0}^{1}{{{\left[ f\left( x \right) \right]}^{3}}dx}=3\left( {{x}^{2}}+9x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=30$
$\Rightarrow \int\limits_{0}^{1}{\left[ {f}'\left( x \right).{{\left[ f\left( x \right) \right]}^{2}}-4\sqrt{{f}'\left( x \right)}.f\left( x \right)+4 \right]dx}=0\Rightarrow \int\limits_{0}^{1}{{{\left[ \sqrt{{f}'\left( x \right)}.f\left( x \right)-2 \right]}^{2}}dx}=0$
$\Rightarrow \sqrt{{f}'\left( x \right)}.f\left( x \right)-2=0\Rightarrow {{\left[ f\left( x \right) \right]}^{2}}.{f}'\left( x \right)=2\Rightarrow \int\limits_{{}}^{{}}{{{\left[ f\left( x \right) \right]}^{2}}.{f}'\left( x \right)dx}=2x+{{C}_{1}}$
$\Rightarrow \int\limits_{{}}^{{}}{{{\left[ f\left( x \right) \right]}^{2}}d\left[ f\left( x \right) \right]}=2x+{{C}_{1}}\Rightarrow \dfrac{{{\left[ f\left( x \right) \right]}^{3}}}{3}=2x+{{C}_{2}}$
Mà $f\left( 0 \right)=3\Rightarrow {{C}_{2}}=9\Rightarrow {{\left[ f\left( x \right) \right]}^{3}}=3\left( 2x+9 \right)\Rightarrow \int\limits_{0}^{1}{{{\left[ f\left( x \right) \right]}^{3}}dx}=3\left( {{x}^{2}}+9x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=30$
Đáp án A.