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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên đoạn $\left[ 0;\dfrac{\pi }{3} \right]$.
Biết ${f}'\left( x \right).\cos x+f\left( x \right).\sin x=1,\forall x\in \left[ 0;\dfrac{\pi }{3} \right]$ và $f\left( 0 \right)=1$. Tính tích phân $I=\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)dx.}$
A. $I=\dfrac{\sqrt{3}+1}{2}.$
B. $I=\dfrac{\sqrt{3}-1}{2}.$
C. $I=\dfrac{1}{2}.$
D. $I=\dfrac{1}{2}+\dfrac{\pi }{3}.$
Biết ${f}'\left( x \right).\cos x+f\left( x \right).\sin x=1,\forall x\in \left[ 0;\dfrac{\pi }{3} \right]$ và $f\left( 0 \right)=1$. Tính tích phân $I=\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)dx.}$
A. $I=\dfrac{\sqrt{3}+1}{2}.$
B. $I=\dfrac{\sqrt{3}-1}{2}.$
C. $I=\dfrac{1}{2}.$
D. $I=\dfrac{1}{2}+\dfrac{\pi }{3}.$
Xét trên đoạn $\left[ 0;\dfrac{\pi }{3} \right]$, ta có
${f}'\left( x \right).\cos x+f\left( x \right).\sin x=1\Leftrightarrow \dfrac{{f}'\left( x \right).\cos x+f\left( x \right).\sin x}{{{\cos }^{2}}x}=\dfrac{1}{{{\cos }^{2}}x}$
$\Rightarrow \dfrac{{f}'\left( x \right).\cos x+f\left( x \right).\sin x}{{{\cos }^{2}}x}-\dfrac{1}{{{\cos }^{2}}x}=0\Rightarrow \left[ \dfrac{f\left( x \right)}{\cos x}-\tan x \right]=0\Rightarrow \dfrac{f\left( x \right)}{\cos x}-\tan x=C$
Mà $f\left( 0 \right)=1$ suy ra $C=1.$ Suy ra $\Rightarrow f\left( x \right)=\sin x+\cos x.$
Vậy $I=\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}dx=\int\limits_{0}^{\dfrac{\pi }{3}}{\left( \sin x+\cos x \right)dx=\left( \sin x-\cos x \right)\left| _{\begin{smallmatrix}
\\
0
\end{smallmatrix}}^{\dfrac{\pi }{3}} \right.=\dfrac{\sqrt{3}+1}{2}.}$
${f}'\left( x \right).\cos x+f\left( x \right).\sin x=1\Leftrightarrow \dfrac{{f}'\left( x \right).\cos x+f\left( x \right).\sin x}{{{\cos }^{2}}x}=\dfrac{1}{{{\cos }^{2}}x}$
$\Rightarrow \dfrac{{f}'\left( x \right).\cos x+f\left( x \right).\sin x}{{{\cos }^{2}}x}-\dfrac{1}{{{\cos }^{2}}x}=0\Rightarrow \left[ \dfrac{f\left( x \right)}{\cos x}-\tan x \right]=0\Rightarrow \dfrac{f\left( x \right)}{\cos x}-\tan x=C$
Mà $f\left( 0 \right)=1$ suy ra $C=1.$ Suy ra $\Rightarrow f\left( x \right)=\sin x+\cos x.$
Vậy $I=\int\limits_{0}^{\dfrac{\pi }{3}}{f\left( x \right)}dx=\int\limits_{0}^{\dfrac{\pi }{3}}{\left( \sin x+\cos x \right)dx=\left( \sin x-\cos x \right)\left| _{\begin{smallmatrix}
\\
0
\end{smallmatrix}}^{\dfrac{\pi }{3}} \right.=\dfrac{\sqrt{3}+1}{2}.}$
Đáp án A.