Google
Câu hỏi: Cho hàm số $f\left( x \right)=\left\{ \begin{aligned}
& x+1 khi x\le 2 \\
& {{x}^{2}}-1 khi x>2 \\
\end{aligned} \right. $. Giá trị của tích phân $ \int\limits_{0}^{2\sqrt{2}}{\dfrac{2xf\left( \sqrt{1+{{x}^{2}}} \right)}{\sqrt{1+{{x}^{2}}}}}\text{d}x$ bằng
A. $\dfrac{47}{3}$.
B. $\dfrac{79}{3}$.
C. $\dfrac{79}{6}$.
D. $\dfrac{47}{6}$.
& x+1 khi x\le 2 \\
& {{x}^{2}}-1 khi x>2 \\
\end{aligned} \right. $. Giá trị của tích phân $ \int\limits_{0}^{2\sqrt{2}}{\dfrac{2xf\left( \sqrt{1+{{x}^{2}}} \right)}{\sqrt{1+{{x}^{2}}}}}\text{d}x$ bằng
A. $\dfrac{47}{3}$.
B. $\dfrac{79}{3}$.
C. $\dfrac{79}{6}$.
D. $\dfrac{47}{6}$.
Xét $I=\int\limits_{0}^{2\sqrt{2}}{\dfrac{2xf\left( \sqrt{1+{{x}^{2}}} \right)}{\sqrt{1+{{x}^{2}}}}}\text{d}x$.
Đặt $t=\sqrt{1+{{x}^{2}}}\Rightarrow xdx=tdt$ $x=0\Rightarrow t=1; x=2\sqrt{2}\Rightarrow t=3$
$\Rightarrow I=\int\limits_{1}^{3}{2t\dfrac{f\left( t \right)}{t}\text{d}t}=2\left[ \int\limits_{1}^{2}{f\left( t \right)\text{d}t}+\int\limits_{2}^{3}{f\left( t \right)\text{d}t} \right]=2\left[ \int\limits_{1}^{2}{\left( t+1 \right)\text{d}t}+\int\limits_{2}^{3}{\left( {{t}^{2}}-1 \right)\text{d}t} \right]=\dfrac{47}{3}$.
Đặt $t=\sqrt{1+{{x}^{2}}}\Rightarrow xdx=tdt$ $x=0\Rightarrow t=1; x=2\sqrt{2}\Rightarrow t=3$
$\Rightarrow I=\int\limits_{1}^{3}{2t\dfrac{f\left( t \right)}{t}\text{d}t}=2\left[ \int\limits_{1}^{2}{f\left( t \right)\text{d}t}+\int\limits_{2}^{3}{f\left( t \right)\text{d}t} \right]=2\left[ \int\limits_{1}^{2}{\left( t+1 \right)\text{d}t}+\int\limits_{2}^{3}{\left( {{t}^{2}}-1 \right)\text{d}t} \right]=\dfrac{47}{3}$.
Đáp án A.
