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Câu hỏi: Tích phân $\int\limits_{0}^{2}{4048{{\left( 2x-1 \right)}^{2023}}\text{d}x}$ bằng
A. ${{2.3}^{2024}}-2$
B. ${{3}^{2024}}-1$
C. ${{3}^{2024}}+1$
D. ${{2.3}^{2024}}+2$
A. ${{2.3}^{2024}}-2$
B. ${{3}^{2024}}-1$
C. ${{3}^{2024}}+1$
D. ${{2.3}^{2024}}+2$
Đặt $t=2x-1\Rightarrow \text{d}t=2\text{d}x$
Đổi cận $\left\{ \begin{aligned}
& x=0\Rightarrow t=-1 \\
& x=2\Rightarrow t=3 \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{-1}^{3}{2024{{t}^{2023}}\text{d}t}={{t}^{2024}}\left| \begin{aligned}
& 3 \\
& -1 \\
\end{aligned} \right.={{3}^{2024}}-1$.
Đổi cận $\left\{ \begin{aligned}
& x=0\Rightarrow t=-1 \\
& x=2\Rightarrow t=3 \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{-1}^{3}{2024{{t}^{2023}}\text{d}t}={{t}^{2024}}\left| \begin{aligned}
& 3 \\
& -1 \\
\end{aligned} \right.={{3}^{2024}}-1$.
Đáp án B.