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