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Câu hỏi: Cho hàm số $y=f(x)$ có đạo hàm là ${f}'(x)={{x}^{2021}}+2022,\forall x\in \mathbb{R}$ và $f(1)=1011$. Giá trị của $\int\limits_{0}^{2}{f\left( \dfrac{x}{2} \right)\text{d}x}$ bằng
A. $-\dfrac{1}{2023}$
B. $-\dfrac{1}{4046}$
C. $-\dfrac{2}{2023}$
D. $\dfrac{1}{2023}$
A. $-\dfrac{1}{2023}$
B. $-\dfrac{1}{4046}$
C. $-\dfrac{2}{2023}$
D. $\dfrac{1}{2023}$
Đặt $t=\dfrac{x}{2}\Rightarrow \text{d}x=2\text{d}t$. Đổi cận: $x=0\to t=0$ ; $x=2\to t=1$.
Ta có: $I=\int\limits_{0}^{2}{f\left( \dfrac{x}{2} \right)\text{d}x}=2\int\limits_{0}^{1}{f(t)\text{d}t}$.
Đặt $\left\{ \begin{aligned}
& u=f(t) \\
& \text{d}v=\text{d}t \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={f}'(t)\text{d}t \\
& v=t \\
\end{aligned} \right.$. Khi đó:
$\int\limits_{0}^{1}{f(t)\text{d}t}=\left. t.f(t) \right|_{0}^{1}-\int\limits_{0}^{1}{t.{f}'(t)\text{d}t}=f(1)-\int\limits_{0}^{1}{t\left( {{t}^{2021}}+2022 \right)\text{d}t}=1011-\left( \dfrac{1}{2023}+1011 \right)=-\dfrac{1}{2023}$.
Suy ra: $I=-\dfrac{2}{2023}$.
Ta có: $I=\int\limits_{0}^{2}{f\left( \dfrac{x}{2} \right)\text{d}x}=2\int\limits_{0}^{1}{f(t)\text{d}t}$.
Đặt $\left\{ \begin{aligned}
& u=f(t) \\
& \text{d}v=\text{d}t \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={f}'(t)\text{d}t \\
& v=t \\
\end{aligned} \right.$. Khi đó:
$\int\limits_{0}^{1}{f(t)\text{d}t}=\left. t.f(t) \right|_{0}^{1}-\int\limits_{0}^{1}{t.{f}'(t)\text{d}t}=f(1)-\int\limits_{0}^{1}{t\left( {{t}^{2021}}+2022 \right)\text{d}t}=1011-\left( \dfrac{1}{2023}+1011 \right)=-\dfrac{1}{2023}$.
Suy ra: $I=-\dfrac{2}{2023}$.
Đáp án C.