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Câu hỏi: Cho hàm số ch phân $y=f\left( x \right)$ có đạo hàm liên tục trên $\mathbb{R}$ thỏa mãn $f\left( x \right)+f\left( \dfrac{\pi }{2}-x \right)=1+\sin 2x$, với mọi $x\in \mathbb{R}$ và $f\left( 0 \right)=0$. Giá trị của tích phân $\int\limits_{0}^{\dfrac{\pi }{4}}{x.{f}'\left( 2x \right)\text{d}x}$ gần nhất với giá trị nào trong các giá trị sau?
A. $1,08$.
B. $0,07$.
C. $0,83$.
D. $0,17$.
A. $1,08$.
B. $0,07$.
C. $0,83$.
D. $0,17$.
Xét $A=\int\limits_{0}^{\dfrac{\pi }{4}}{x.{f}'\left( 2x \right)\text{d}x}$
+) Đặt $t=2x\Rightarrow dx=\dfrac{1}{2}dt$
+) $x=0\Rightarrow t=0; x=\dfrac{\pi }{4}\Rightarrow t=\dfrac{\pi }{2}$
$\Rightarrow A=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{t}{2}.{f}'\left( t \right)\text{d}t}=\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{t.{f}'\left( t \right)\text{d}t}=\dfrac{1}{4}\left[ t\left. f\left( t \right) \right|_{0}^{\dfrac{\pi }{2}}-\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( t \right)\text{d}t} \right]=\dfrac{1}{4}\left[ \dfrac{\pi }{2}f\left( \dfrac{\pi }{2} \right)-\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( t \right)\text{d}t} \right]$
Xét $f\left( x \right)+f\left( \dfrac{\pi }{2}-x \right)=1+\sin 2x$ (*)
+) Từ (*) suy ra $f\left( 0 \right)+f\left( \dfrac{\pi }{2} \right)=1$, mà $f\left( 0 \right)=0$ $\Rightarrow f\left( \dfrac{\pi }{2} \right)=1$
+) Từ (*) suy ra $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}+\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+\sin 2x \right)\text{d}x}$ (**)
Đặt $I=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)\text{d}x}=I$
Khi đó (**) $\Leftrightarrow 2I=\dfrac{\pi }{2}+1\Leftrightarrow I=\dfrac{\pi }{4}+\dfrac{1}{2}$
$\Rightarrow A=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{t}{2}.{f}'\left( t \right)\text{d}t}=\dfrac{1}{4}\left[ \dfrac{\pi }{2}.1-\dfrac{\pi }{4}-\dfrac{1}{2} \right]=\dfrac{\pi }{16}-\dfrac{1}{8}\approx 0,07$.
+) Đặt $t=2x\Rightarrow dx=\dfrac{1}{2}dt$
+) $x=0\Rightarrow t=0; x=\dfrac{\pi }{4}\Rightarrow t=\dfrac{\pi }{2}$
$\Rightarrow A=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{t}{2}.{f}'\left( t \right)\text{d}t}=\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{t.{f}'\left( t \right)\text{d}t}=\dfrac{1}{4}\left[ t\left. f\left( t \right) \right|_{0}^{\dfrac{\pi }{2}}-\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( t \right)\text{d}t} \right]=\dfrac{1}{4}\left[ \dfrac{\pi }{2}f\left( \dfrac{\pi }{2} \right)-\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( t \right)\text{d}t} \right]$
Xét $f\left( x \right)+f\left( \dfrac{\pi }{2}-x \right)=1+\sin 2x$ (*)
+) Từ (*) suy ra $f\left( 0 \right)+f\left( \dfrac{\pi }{2} \right)=1$, mà $f\left( 0 \right)=0$ $\Rightarrow f\left( \dfrac{\pi }{2} \right)=1$
+) Từ (*) suy ra $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}+\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+\sin 2x \right)\text{d}x}$ (**)
Đặt $I=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)\text{d}x}=I$
Khi đó (**) $\Leftrightarrow 2I=\dfrac{\pi }{2}+1\Leftrightarrow I=\dfrac{\pi }{4}+\dfrac{1}{2}$
$\Rightarrow A=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{t}{2}.{f}'\left( t \right)\text{d}t}=\dfrac{1}{4}\left[ \dfrac{\pi }{2}.1-\dfrac{\pi }{4}-\dfrac{1}{2} \right]=\dfrac{\pi }{16}-\dfrac{1}{8}\approx 0,07$.
Đáp án B.
