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Câu hỏi: Cho hàm số f(x) 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. 33.
B. 10.
C. 21.
D. 19.
A. 33.
B. 10.
C. 21.
D. 19.
Ta có $\int\limits_{0}^{1}{\left[ {f}'\left( x \right).{{\left[ f\left( x \right) \right]}^{2}}+4 \right]dx-4\int\limits_{0}^{1}{\sqrt{{f}'\left( x \right)}.f\left( x \right)dx}=0}$
$\Rightarrow \int\limits_{0}^{1}{\left[ {f}'\left( x \right).{{\left[ f\left( x \right) \right]}^{2}}-4\sqrt{{f}'\left( x \right)}.f\left( x \right)+4 \right]dx}=0\Rightarrow \int\limits_{0}^{1}{{{\left[ \sqrt{{f}'\left( x \right)}.f\left( x \right)-2 \right]}^{2}}dx}=0$
$\Rightarrow \sqrt{{f}'\left( x \right)}.f\left( x \right)-2=0\Rightarrow {{\left[ f\left( x \right) \right]}^{2}}.{f}'\left( x \right)=4\Rightarrow \int{{{\left[ f\left( x \right) \right]}^{2}}.{f}'\left( x \right)dx}=4x+{{C}_{1}}$
$\Rightarrow \int{{{\left[ f\left( x \right) \right]}^{2}}d\left[ f\left( x \right) \right]=4x+{{C}_{1}}\Rightarrow \dfrac{{{\left[ f\left( x \right) \right]}^{3}}}{3}=4x+{{C}_{2}}}$.
Mà $f\left( 0 \right)=3\Rightarrow {{C}_{2}}=9\Rightarrow {{\left[ f\left( x \right) \right]}^{3}}=3\left( 4x+9 \right)\Rightarrow \int\limits_{0}^{1}{{{\left[ f\left( x \right) \right]}^{3}}dx}=\left. 3\left( 2{{x}^{2}}+9x \right) \right|_{0}^{1}=33$.
$\Rightarrow \int\limits_{0}^{1}{\left[ {f}'\left( x \right).{{\left[ f\left( x \right) \right]}^{2}}-4\sqrt{{f}'\left( x \right)}.f\left( x \right)+4 \right]dx}=0\Rightarrow \int\limits_{0}^{1}{{{\left[ \sqrt{{f}'\left( x \right)}.f\left( x \right)-2 \right]}^{2}}dx}=0$
$\Rightarrow \sqrt{{f}'\left( x \right)}.f\left( x \right)-2=0\Rightarrow {{\left[ f\left( x \right) \right]}^{2}}.{f}'\left( x \right)=4\Rightarrow \int{{{\left[ f\left( x \right) \right]}^{2}}.{f}'\left( x \right)dx}=4x+{{C}_{1}}$
$\Rightarrow \int{{{\left[ f\left( x \right) \right]}^{2}}d\left[ f\left( x \right) \right]=4x+{{C}_{1}}\Rightarrow \dfrac{{{\left[ f\left( x \right) \right]}^{3}}}{3}=4x+{{C}_{2}}}$.
Mà $f\left( 0 \right)=3\Rightarrow {{C}_{2}}=9\Rightarrow {{\left[ f\left( x \right) \right]}^{3}}=3\left( 4x+9 \right)\Rightarrow \int\limits_{0}^{1}{{{\left[ f\left( x \right) \right]}^{3}}dx}=\left. 3\left( 2{{x}^{2}}+9x \right) \right|_{0}^{1}=33$.
Đáp án A.