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Câu hỏi: Cho hàm số $y=f\left( x \right)=\left\{ \begin{aligned}
& {{x}^{2}}+3x-1\text{ khi }x\le 1 \\
& 5-2x\text{ khi }x>1 \\
\end{aligned} \right. $. Tính tích phân:$ I=\int\limits_{0}^{\pi }{\sin x.f(\cos x)\text{d}x+\int\limits_{0}^{1}{f(3-2x)\text{d}x}}$
A. $\dfrac{58}{3}$.
B. $\dfrac{8}{3}$.
C. $\dfrac{2}{3}$.
D. $-\dfrac{1}{3}$.
& {{x}^{2}}+3x-1\text{ khi }x\le 1 \\
& 5-2x\text{ khi }x>1 \\
\end{aligned} \right. $. Tính tích phân:$ I=\int\limits_{0}^{\pi }{\sin x.f(\cos x)\text{d}x+\int\limits_{0}^{1}{f(3-2x)\text{d}x}}$
A. $\dfrac{58}{3}$.
B. $\dfrac{8}{3}$.
C. $\dfrac{2}{3}$.
D. $-\dfrac{1}{3}$.
Ta có: $\underset{x\to {{1}^{-}}}{\mathop{\lim }} \left( {{x}^{2}}+3x-1 \right)=3;\underset{x\to {{1}^{+}}}{\mathop{\lim }} \left( 5-2x \right)=3;f(1)=3$.
$\Rightarrow \underset{x\to {{1}^{+}}}{\mathop{\lim }} f(x)=\underset{x\to {{1}^{-}}}{\mathop{\lim }} f(x)=f(1)$ $\Rightarrow f(x)$ liên tục tại $x=1$.
$I=\int\limits_{0}^{\pi }{\sin x.f(\cos x)\text{d}x+\int\limits_{0}^{1}{f(3-2x)\text{d}x}}={{I}_{1}}+{{I}_{2}}$.
${{I}_{1}}=\int\limits_{0}^{\pi }{\sin x.f(\cos x)\text{d}x}$ ; đặt $t=\cos x\Rightarrow \text{d}t=-\sin x\text{d}x$.
Khi đó: ${{I}_{1}}=\int\limits_{1}^{-1}{-f(t)\text{d}t=\int\limits_{-1}^{1}{\left( {{x}^{2}}+3x-1 \right)\text{d}x=-\dfrac{4}{3}}}$.
${{I}_{2}}=\int\limits_{0}^{1}{f(3-2x)\text{d}x}$ ; đặt $u=3-2x\Rightarrow \text{d}u=-2\text{d}x$.
Khi đó: ${{I}_{2}}=\int\limits_{3}^{1}{-\dfrac{1}{2}f(u)\text{d}u=\dfrac{1}{2}}\int\limits_{1}^{3}{(5-2x)\text{d}x=1}$.
Vậy $I=-\dfrac{4}{3}+1=-\dfrac{1}{3}$.
$\Rightarrow \underset{x\to {{1}^{+}}}{\mathop{\lim }} f(x)=\underset{x\to {{1}^{-}}}{\mathop{\lim }} f(x)=f(1)$ $\Rightarrow f(x)$ liên tục tại $x=1$.
$I=\int\limits_{0}^{\pi }{\sin x.f(\cos x)\text{d}x+\int\limits_{0}^{1}{f(3-2x)\text{d}x}}={{I}_{1}}+{{I}_{2}}$.
${{I}_{1}}=\int\limits_{0}^{\pi }{\sin x.f(\cos x)\text{d}x}$ ; đặt $t=\cos x\Rightarrow \text{d}t=-\sin x\text{d}x$.
Khi đó: ${{I}_{1}}=\int\limits_{1}^{-1}{-f(t)\text{d}t=\int\limits_{-1}^{1}{\left( {{x}^{2}}+3x-1 \right)\text{d}x=-\dfrac{4}{3}}}$.
${{I}_{2}}=\int\limits_{0}^{1}{f(3-2x)\text{d}x}$ ; đặt $u=3-2x\Rightarrow \text{d}u=-2\text{d}x$.
Khi đó: ${{I}_{2}}=\int\limits_{3}^{1}{-\dfrac{1}{2}f(u)\text{d}u=\dfrac{1}{2}}\int\limits_{1}^{3}{(5-2x)\text{d}x=1}$.
Vậy $I=-\dfrac{4}{3}+1=-\dfrac{1}{3}$.
Đáp án D.