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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm và liên tục trên $\left[ 0;1 \right]$, thỏa mãn $\int\limits_{1}^{2}{f\left( x-1 \right)dx}=3$ và $f\left( 1 \right)=4$. Tích phân $\int\limits_{0}^{1}{{{x}^{3}}{f}'\left( {{x}^{2}} \right)dx}$ bằng
A. $-1$
B. $-\dfrac{1}{2}$.
C. $\dfrac{1}{2}$.
D. 1.
A. $-1$
B. $-\dfrac{1}{2}$.
C. $\dfrac{1}{2}$.
D. 1.
Đặt $t={{x}^{2}}\Rightarrow dt=2xdx$. Đổi cận $\left[ \begin{aligned}
& x=0\Rightarrow t=0 \\
& x=1\Rightarrow t=1 \\
\end{aligned} \right. $ ta có: $ \int\limits_{0}^{1}{{{x}^{3}}{f}'\left( {{x}^{2}} \right)dx}=\dfrac{1}{2}\int\limits_{0}^{1}{t.{f}'\left( t \right)dt}$
$=\dfrac{1}{2}\int\limits_{0}^{1}{t.d\left[ f\left( t \right) \right]}=\dfrac{1}{2}\left[ \left. t.f\left( t \right) \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( t \right)dt} \right]=\dfrac{1}{2}\left[ f\left( 1 \right)-\int\limits_{0}^{1}{f\left( x \right)dx} \right]$
Lại có: $\int\limits_{1}^{2}{f\left( x-1 \right)dx}=\int\limits_{1}^{2}{f\left( x-1 \right)d\left( x-1 \right)}\xrightarrow[{}]{u=x-1}\int\limits_{1}^{2}{f\left( x-1 \right)dx}=\int\limits_{0}^{1}{f\left( u \right)du}=\int\limits_{0}^{1}{f\left( x \right)dx}=3$
Suy ra $\int\limits_{0}^{1}{{{x}^{3}}{f}'\left( {{x}^{2}} \right)dx}=\dfrac{1}{2}\left( 4-3 \right)=\dfrac{1}{2}$.
& x=0\Rightarrow t=0 \\
& x=1\Rightarrow t=1 \\
\end{aligned} \right. $ ta có: $ \int\limits_{0}^{1}{{{x}^{3}}{f}'\left( {{x}^{2}} \right)dx}=\dfrac{1}{2}\int\limits_{0}^{1}{t.{f}'\left( t \right)dt}$
$=\dfrac{1}{2}\int\limits_{0}^{1}{t.d\left[ f\left( t \right) \right]}=\dfrac{1}{2}\left[ \left. t.f\left( t \right) \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( t \right)dt} \right]=\dfrac{1}{2}\left[ f\left( 1 \right)-\int\limits_{0}^{1}{f\left( x \right)dx} \right]$
Lại có: $\int\limits_{1}^{2}{f\left( x-1 \right)dx}=\int\limits_{1}^{2}{f\left( x-1 \right)d\left( x-1 \right)}\xrightarrow[{}]{u=x-1}\int\limits_{1}^{2}{f\left( x-1 \right)dx}=\int\limits_{0}^{1}{f\left( u \right)du}=\int\limits_{0}^{1}{f\left( x \right)dx}=3$
Suy ra $\int\limits_{0}^{1}{{{x}^{3}}{f}'\left( {{x}^{2}} \right)dx}=\dfrac{1}{2}\left( 4-3 \right)=\dfrac{1}{2}$.
Đáp án C.