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Câu hỏi: Cho hàm số $f(x)=\left\{\begin{array}{ll}2 x-4 & \text { khi } x \geq 4 \\ \dfrac{1}{4} x^3-x^2+x & \text { khi } x<4\end{array}\right.$. Tích phân $\int_0^{\dfrac{\pi}{2}} f\left(2 \sin ^2 x+3\right) \sin 2 x \mathrm{~d} x$ bằng
A. $\dfrac{341}{48}$.
B. $\dfrac{341}{96}$.
C. $\dfrac{28}{3}$.
D. 8 .
A. $\dfrac{341}{48}$.
B. $\dfrac{341}{96}$.
C. $\dfrac{28}{3}$.
D. 8 .
Ta có
$
\begin{aligned}
& \lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{+}}(2 x-4)=4 ; \lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{-}}\left(\dfrac{1}{4} x^3-x^2+x\right)=4 ; f(4)=4 \\
& \Rightarrow \lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{-}} f(x)=f(4)
\end{aligned}
$
Nên hàm số đã cho liên tục tại $x=4$
Xét $I=\int_0^{\dfrac{\pi}{2}} f\left(2 \sin ^2 x+3\right) \sin 2 x \mathrm{~d} x$
Đặt $2 \sin ^2 x+3=t \Rightarrow \sin 2 x \mathrm{~d} x=\dfrac{1}{2} \mathrm{~d} t$
Với $x=0 \Rightarrow t=3$
$x=\dfrac{\pi}{2} \Rightarrow t=5$
$\Rightarrow I=\int_3^5 f(t) \dfrac{1}{2} \mathrm{~d} t=\dfrac{1}{2} \int_3^5 f(t) \mathrm{d} t=\dfrac{1}{2} \int_3^4\left(\dfrac{1}{4} t^3-t^2+t\right) \mathrm{d} t+\dfrac{1}{2} \int_4^5(2 t-4) \mathrm{d} t=\dfrac{341}{96}$.
$
\begin{aligned}
& \lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{+}}(2 x-4)=4 ; \lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{-}}\left(\dfrac{1}{4} x^3-x^2+x\right)=4 ; f(4)=4 \\
& \Rightarrow \lim _{x \rightarrow 4^{+}} f(x)=\lim _{x \rightarrow 4^{-}} f(x)=f(4)
\end{aligned}
$
Nên hàm số đã cho liên tục tại $x=4$
Xét $I=\int_0^{\dfrac{\pi}{2}} f\left(2 \sin ^2 x+3\right) \sin 2 x \mathrm{~d} x$
Đặt $2 \sin ^2 x+3=t \Rightarrow \sin 2 x \mathrm{~d} x=\dfrac{1}{2} \mathrm{~d} t$
Với $x=0 \Rightarrow t=3$
$x=\dfrac{\pi}{2} \Rightarrow t=5$
$\Rightarrow I=\int_3^5 f(t) \dfrac{1}{2} \mathrm{~d} t=\dfrac{1}{2} \int_3^5 f(t) \mathrm{d} t=\dfrac{1}{2} \int_3^4\left(\dfrac{1}{4} t^3-t^2+t\right) \mathrm{d} t+\dfrac{1}{2} \int_4^5(2 t-4) \mathrm{d} t=\dfrac{341}{96}$.
Đáp án B.