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Câu hỏi: Cho $I=\int\limits_{1}^{5}{f\left( x \right)dx}=26$. Khi đó $J=\int\limits_{0}^{2}{x.\left[ f\left( {{x}^{2}}+1 \right)+1 \right]dx}$ bằng
A. 13.
B. 52.
C. 54.
D. 15.
A. 13.
B. 52.
C. 54.
D. 15.
Đặt $u={{x}^{2}}+1\Rightarrow du=2xdx$.
Khi đó $J=\dfrac{1}{2}\int\limits_{1}^{5}{\left[ f\left( u \right)+1 \right]du}=\dfrac{1}{2}\left( \int\limits_{1}^{5}{f\left( u \right)du}+\int\limits_{1}^{5}{1du} \right)=\dfrac{1}{2}\left( 26+x\left| \begin{aligned}
& ^{5} \\
& _{1} \\
\end{aligned} \right. \right)=\dfrac{1}{2}\left( 26+5-1 \right)=15$.
Khi đó $J=\dfrac{1}{2}\int\limits_{1}^{5}{\left[ f\left( u \right)+1 \right]du}=\dfrac{1}{2}\left( \int\limits_{1}^{5}{f\left( u \right)du}+\int\limits_{1}^{5}{1du} \right)=\dfrac{1}{2}\left( 26+x\left| \begin{aligned}
& ^{5} \\
& _{1} \\
\end{aligned} \right. \right)=\dfrac{1}{2}\left( 26+5-1 \right)=15$.
Đáp án D.