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Câu hỏi: Cho hàm số $y=f\left( x \right)=\left\{ \begin{aligned}
& {{x}^{2}}+3,x\ge 1 \\
& 5-x,x<1 \\
\end{aligned} \right. $. Tính$ I=2\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \sin x \right)}\cos x~\text{d}x+\dfrac{3}{2}\int\limits_{0}^{1}{f\left( 3-2x \right)}~\text{d}x$.
A. $I=\dfrac{32}{3}$.
B. $I=20$.
C. $I=32$.
D. $I=31$.
& {{x}^{2}}+3,x\ge 1 \\
& 5-x,x<1 \\
\end{aligned} \right. $. Tính$ I=2\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \sin x \right)}\cos x~\text{d}x+\dfrac{3}{2}\int\limits_{0}^{1}{f\left( 3-2x \right)}~\text{d}x$.
A. $I=\dfrac{32}{3}$.
B. $I=20$.
C. $I=32$.
D. $I=31$.
Ta có: $I=2\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \sin x \right)}\cos x~\text{d}x+\dfrac{3}{2}\int\limits_{0}^{1}{f\left( 3-2x \right)}~\text{d}x=2{{I}_{1}}+\dfrac{3}{2}{{I}_{2}}$.
+ ${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \sin x \right)}\cos x~\text{d}x=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \sin x \right)}~\text{d}\left( \sin x \right)=\int\limits_{0}^{1}{f\left( x \right)}~\text{d}x$.
$\Rightarrow $ ${{I}_{1}}=\int\limits_{0}^{1}{\left( 5-x \right)}~\text{d}x=\left. \left( 5x-\dfrac{{{x}^{2}}}{2} \right) \right|_{0}^{1}=\dfrac{9}{2}$.
+ ${{I}_{2}}=\int\limits_{0}^{1}{f\left( 3-2x \right)}~\text{d}x=\dfrac{-1}{2}\int\limits_{0}^{1}{f\left( 3-2x \right)}~\text{d}\left( 3-2x \right)=\dfrac{-1}{2}\int\limits_{3}^{1}{f\left( x \right)}~\text{d}x=\dfrac{1}{2}\int\limits_{1}^{3}{f\left( x \right)}~\text{d}x$.
$\Rightarrow $ ${{I}_{2}}=\dfrac{1}{2}\int\limits_{1}^{3}{f\left( x \right)}~\text{d}x=\dfrac{1}{2}\int\limits_{1}^{3}{\left( {{x}^{2}}+3 \right)}~\text{d}x=\dfrac{1}{2}\left. \left( \dfrac{{{x}^{3}}}{3}+3x \right) \right|_{1}^{3}=\dfrac{22}{3}$.
Vậy $I=2{{I}_{1}}+\dfrac{3}{2}{{I}_{2}}=2.\dfrac{9}{2}+\dfrac{3}{2}.\dfrac{22}{3}=20$.
+ ${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \sin x \right)}\cos x~\text{d}x=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \sin x \right)}~\text{d}\left( \sin x \right)=\int\limits_{0}^{1}{f\left( x \right)}~\text{d}x$.
$\Rightarrow $ ${{I}_{1}}=\int\limits_{0}^{1}{\left( 5-x \right)}~\text{d}x=\left. \left( 5x-\dfrac{{{x}^{2}}}{2} \right) \right|_{0}^{1}=\dfrac{9}{2}$.
+ ${{I}_{2}}=\int\limits_{0}^{1}{f\left( 3-2x \right)}~\text{d}x=\dfrac{-1}{2}\int\limits_{0}^{1}{f\left( 3-2x \right)}~\text{d}\left( 3-2x \right)=\dfrac{-1}{2}\int\limits_{3}^{1}{f\left( x \right)}~\text{d}x=\dfrac{1}{2}\int\limits_{1}^{3}{f\left( x \right)}~\text{d}x$.
$\Rightarrow $ ${{I}_{2}}=\dfrac{1}{2}\int\limits_{1}^{3}{f\left( x \right)}~\text{d}x=\dfrac{1}{2}\int\limits_{1}^{3}{\left( {{x}^{2}}+3 \right)}~\text{d}x=\dfrac{1}{2}\left. \left( \dfrac{{{x}^{3}}}{3}+3x \right) \right|_{1}^{3}=\dfrac{22}{3}$.
Vậy $I=2{{I}_{1}}+\dfrac{3}{2}{{I}_{2}}=2.\dfrac{9}{2}+\dfrac{3}{2}.\dfrac{22}{3}=20$.
Đáp án B.