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Câu hỏi: Cho hai hàm số $y=f\left( x \right), y=g\left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 0; 1 \right]$ thỏa mãn $f\left( 1 \right)g\left( 1 \right)=f\left( 0 \right)g\left( 0 \right), {f}'\left( x \right)g\left( x \right)=\sqrt{2x+2}, \forall x\in \left[ 0; 1 \right]$. Tính tích phân $I=\int\limits_{0}^{1}{f\left( x \right).{g}'\left( x \right)dx}$
A. $I=\dfrac{\left( 3\sqrt{2}-4 \right)}{5}$
B. $I=\dfrac{5\left( 3\sqrt{2}-4 \right)}{3}$
C. $I=\dfrac{2\left( \sqrt{2}-4 \right)}{3}$
D. $I=\dfrac{2\left( 4-\sqrt{2} \right)}{7}$
A. $I=\dfrac{\left( 3\sqrt{2}-4 \right)}{5}$
B. $I=\dfrac{5\left( 3\sqrt{2}-4 \right)}{3}$
C. $I=\dfrac{2\left( \sqrt{2}-4 \right)}{3}$
D. $I=\dfrac{2\left( 4-\sqrt{2} \right)}{7}$
Ta có ${f}'\left( x \right)g\left( x \right)=\sqrt{2x+2}, \forall x\in \left[ 0; 1 \right]$
Lấy tích phân 2 vế trên đoạn $\left[ 0; 1 \right]$ ta được: $\int\limits_{0}^{1}{{f}'\left( x \right)g\left( x \right)dx}=\int\limits_{0}^{1}{\sqrt{2x+2}dx}$
Xét tích phân: $A=\int\limits_{0}^{1}{{f}'\left( x \right)g\left( x \right)dx}=\int\limits_{0}^{1}{g\left( x \right)df\left( x \right)}=f\left( x \right)g\left( x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.-\int\limits_{0}^{1}{f\left( x \right){g}'\left( x \right)dx}$
Xét tích phân: $B=\int\limits_{0}^{1}{\sqrt{2x+2}dx}$
Đặt $t=\sqrt{2x+2}\Rightarrow {{t}^{2}}\left( 2x+2 \right)dx\Leftrightarrow 2tdt=2xdx\Leftrightarrow tdt=xdx$
Đổi cận $\left\{ \begin{aligned}
& x=0\Rightarrow t=\sqrt{2} \\
& x=1\Rightarrow t=2 \\
\end{aligned} \right.$
$B=\int\limits_{0}^{1}{\sqrt{2x+2}dx}=\int\limits_{\sqrt{2}}^{2}{{{t}^{2}}dt}=\dfrac{{{t}^{3}}}{3}\left| \begin{aligned}
& ^{2} \\
& _{\sqrt{2}} \\
\end{aligned} \right.=\dfrac{8-2\sqrt{2}}{3}$
Khi đó $I=\int\limits_{0}^{1}{f\left( x \right).{g}'\left( x \right)dx}=f\left( x \right)g\left( x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.-\dfrac{8-2\sqrt{2}}{3}$
$=\left( f\left( 1 \right)g\left( 1 \right)-f\left( 0 \right)g\left( 0 \right) \right)-\dfrac{8-2\sqrt{2}}{3}=\dfrac{2\left( \sqrt{2}-4 \right)}{3}$
Lấy tích phân 2 vế trên đoạn $\left[ 0; 1 \right]$ ta được: $\int\limits_{0}^{1}{{f}'\left( x \right)g\left( x \right)dx}=\int\limits_{0}^{1}{\sqrt{2x+2}dx}$
Xét tích phân: $A=\int\limits_{0}^{1}{{f}'\left( x \right)g\left( x \right)dx}=\int\limits_{0}^{1}{g\left( x \right)df\left( x \right)}=f\left( x \right)g\left( x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.-\int\limits_{0}^{1}{f\left( x \right){g}'\left( x \right)dx}$
Xét tích phân: $B=\int\limits_{0}^{1}{\sqrt{2x+2}dx}$
Đặt $t=\sqrt{2x+2}\Rightarrow {{t}^{2}}\left( 2x+2 \right)dx\Leftrightarrow 2tdt=2xdx\Leftrightarrow tdt=xdx$
Đổi cận $\left\{ \begin{aligned}
& x=0\Rightarrow t=\sqrt{2} \\
& x=1\Rightarrow t=2 \\
\end{aligned} \right.$
$B=\int\limits_{0}^{1}{\sqrt{2x+2}dx}=\int\limits_{\sqrt{2}}^{2}{{{t}^{2}}dt}=\dfrac{{{t}^{3}}}{3}\left| \begin{aligned}
& ^{2} \\
& _{\sqrt{2}} \\
\end{aligned} \right.=\dfrac{8-2\sqrt{2}}{3}$
Khi đó $I=\int\limits_{0}^{1}{f\left( x \right).{g}'\left( x \right)dx}=f\left( x \right)g\left( x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.-\dfrac{8-2\sqrt{2}}{3}$
$=\left( f\left( 1 \right)g\left( 1 \right)-f\left( 0 \right)g\left( 0 \right) \right)-\dfrac{8-2\sqrt{2}}{3}=\dfrac{2\left( \sqrt{2}-4 \right)}{3}$
Đáp án C.