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Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{aligned}
& -3{{x}^{2}}+5x\text{, khi }x\ge 1 \\
& 5-3x,\text{ khi }x<1 \\
\end{aligned} \right.$.
Tính tích phân $I=3\int\limits_{0}^{\dfrac{\pi }{2}}{\cos xf\left( \sin x \right)\text{d}x}+2\int\limits_{0}^{1}{f\left( 3-2x \right)\text{d}x}$.
A. $\dfrac{1}{2}$.
B. $\dfrac{9}{2}$.
C. $\dfrac{11}{2}$.
D. $\dfrac{13}{2}$.
& -3{{x}^{2}}+5x\text{, khi }x\ge 1 \\
& 5-3x,\text{ khi }x<1 \\
\end{aligned} \right.$.
Tính tích phân $I=3\int\limits_{0}^{\dfrac{\pi }{2}}{\cos xf\left( \sin x \right)\text{d}x}+2\int\limits_{0}^{1}{f\left( 3-2x \right)\text{d}x}$.
A. $\dfrac{1}{2}$.
B. $\dfrac{9}{2}$.
C. $\dfrac{11}{2}$.
D. $\dfrac{13}{2}$.
Đặt $t=\sin x$ $\Rightarrow \text{d}t=\cos x\text{d}x$.
Đổi cận: $\left\{ \begin{aligned}
& x=0\Rightarrow t=0 \\
& x=\dfrac{\pi }{2}\Rightarrow t=1 \\
\end{aligned} \right.$.
Khi đó $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos xf\left( \sin x \right)\text{d}x}$ $=\int\limits_{0}^{1}{f\left( t \right)\text{d}t}$ $=\int\limits_{0}^{1}{\left( 5-3t \right)\text{dt}}=\dfrac{7}{2}$.
Đặt $t=3-2x$ $\Rightarrow \text{d}t=-2\text{d}x$.
Đổi cận: $\left\{ \begin{aligned}
& x=0\Rightarrow t=3 \\
& x=1\Rightarrow t=1 \\
\end{aligned} \right.$.
Khi đó $\int\limits_{0}^{1}{f\left( 3-2x \right)\text{d}x}$ $=-\dfrac{1}{2}\int\limits_{3}^{1}{f\left( t \right)\text{d}t}$ $=\dfrac{1}{2}\int\limits_{1}^{3}{\left( -3{{t}^{2}}+5t \right)\text{d}t}=-3$.
Vậy $I=3.\dfrac{7}{2}+2\left( -3 \right)=\dfrac{9}{2}$.
Đổi cận: $\left\{ \begin{aligned}
& x=0\Rightarrow t=0 \\
& x=\dfrac{\pi }{2}\Rightarrow t=1 \\
\end{aligned} \right.$.
Khi đó $\int\limits_{0}^{\dfrac{\pi }{2}}{\cos xf\left( \sin x \right)\text{d}x}$ $=\int\limits_{0}^{1}{f\left( t \right)\text{d}t}$ $=\int\limits_{0}^{1}{\left( 5-3t \right)\text{dt}}=\dfrac{7}{2}$.
Đặt $t=3-2x$ $\Rightarrow \text{d}t=-2\text{d}x$.
Đổi cận: $\left\{ \begin{aligned}
& x=0\Rightarrow t=3 \\
& x=1\Rightarrow t=1 \\
\end{aligned} \right.$.
Khi đó $\int\limits_{0}^{1}{f\left( 3-2x \right)\text{d}x}$ $=-\dfrac{1}{2}\int\limits_{3}^{1}{f\left( t \right)\text{d}t}$ $=\dfrac{1}{2}\int\limits_{1}^{3}{\left( -3{{t}^{2}}+5t \right)\text{d}t}=-3$.
Vậy $I=3.\dfrac{7}{2}+2\left( -3 \right)=\dfrac{9}{2}$.
Đáp án B.