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Câu hỏi: Biết $\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2x+1 \right)\cos x} \text{d}x=a\pi +b$ với $a,b\in \mathbb{Z}$. Giá trị của biểu thức ${{a}^{2}}+{{b}^{2}}$ bằng
A. $1$.
B. $2$.
C. $4$.
D. $0$.
A. $1$.
B. $2$.
C. $4$.
D. $0$.
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2x+1 \right)\cos x} \text{d}x$,
Đặt $\left\{ \begin{aligned}
& u=2x+1 \\
& dv=\cos x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=2\text{d}x \\
& v=\sin x \\
\end{aligned} \right.$ nên:
$I=\left. \left( 2x+1 \right)\sin x \right|_{0}^{\dfrac{\pi }{2}}-2\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x} \text{d}x=\left. \left( 2x+1 \right)\sin x \right|_{0}^{\dfrac{\pi }{2}}+\left. 2\cos x \right|_{0}^{\dfrac{\pi }{2}}=\pi -1\Rightarrow a=1;b=-1\Rightarrow {{a}^{2}}+{{b}^{2}}=2$.
Đặt $\left\{ \begin{aligned}
& u=2x+1 \\
& dv=\cos x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u=2\text{d}x \\
& v=\sin x \\
\end{aligned} \right.$ nên:
$I=\left. \left( 2x+1 \right)\sin x \right|_{0}^{\dfrac{\pi }{2}}-2\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x} \text{d}x=\left. \left( 2x+1 \right)\sin x \right|_{0}^{\dfrac{\pi }{2}}+\left. 2\cos x \right|_{0}^{\dfrac{\pi }{2}}=\pi -1\Rightarrow a=1;b=-1\Rightarrow {{a}^{2}}+{{b}^{2}}=2$.
Đáp án B.