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Câu hỏi: Cho hàm số $f(x)$ có đạo hàm liên tục trên $\left[ 0;1 \right]$, thoả mãn $\int\limits_{-2}^{-1}{f\left( x+2 \right)\text{d}x}=3$ và $f(1)=4.$ Khi đó tích phân $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2xf\prime \left( \sin x \right)\text{d}x}$ bằng
A. $4$.
B. $1$.
C. $2$.
D. $5$.
A. $4$.
B. $1$.
C. $2$.
D. $5$.
$\int\limits_{-2}^{-1}{f\left( x+2 \right)\text{d}x}=3,t=x+2\Rightarrow \int\limits_{0}^{1}{f}(t)\text{d}t=3$.
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2xf\prime \left( \sin x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{2\sin x.\cos x.f\prime \left( \sin x \right)\text{d}x}=\int\limits_{0}^{1}{2}t.f\prime (t)\text{d}t$.
Đặt $\left\{ \begin{aligned}
& u=2t \\
& \text{d}v=f\prime (t)\text{d}t \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=2\text{d}t \\
& v=f(t) \\
\end{aligned} \right.\Rightarrow I=\int\limits_{0}^{1}{2}t.f\prime (t)\text{d}t=\left. 2t.f(t) \right|_{0}^{1}-\int\limits_{0}^{1}{f}(t).2\text{d}t$
$\Rightarrow I=2.f(1)-2\int\limits_{0}^{1}{f}(t)\text{d}t=2\cdot 4-2\cdot 3=2$.
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin 2xf\prime \left( \sin x \right)\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{2}}{2\sin x.\cos x.f\prime \left( \sin x \right)\text{d}x}=\int\limits_{0}^{1}{2}t.f\prime (t)\text{d}t$.
Đặt $\left\{ \begin{aligned}
& u=2t \\
& \text{d}v=f\prime (t)\text{d}t \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=2\text{d}t \\
& v=f(t) \\
\end{aligned} \right.\Rightarrow I=\int\limits_{0}^{1}{2}t.f\prime (t)\text{d}t=\left. 2t.f(t) \right|_{0}^{1}-\int\limits_{0}^{1}{f}(t).2\text{d}t$
$\Rightarrow I=2.f(1)-2\int\limits_{0}^{1}{f}(t)\text{d}t=2\cdot 4-2\cdot 3=2$.
Đáp án C.