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Câu hỏi: Tính tích phân $I=\int\limits_{-2}^{2}{\dfrac{{{x}^{2020}}}{{{e}^{x}}+1}\text{d}x}$
A. $I=\dfrac{{{2}^{2021}}}{2020}$.
B. $I=\dfrac{{{2}^{2021}}}{2021}$.
C. $I=0$.
D. $I=\dfrac{{{2}^{2022}}}{2022}$.
A. $I=\dfrac{{{2}^{2021}}}{2020}$.
B. $I=\dfrac{{{2}^{2021}}}{2021}$.
C. $I=0$.
D. $I=\dfrac{{{2}^{2022}}}{2022}$.
Đặt $t=-x\Rightarrow \text{d}x=-\text{d}t$
Đổi cận
Khi đó ta có: $I=-\int\limits_{2}^{-2}{\dfrac{{{\left( -t \right)}^{2020}}}{{{e}^{-t}}+1}\text{d}t}=\int\limits_{-2}^{2}{\dfrac{{{e}^{t}}.{{t}^{2020}}}{{{e}^{t}}+1}}\text{d}t=\int\limits_{-2}^{2}{\dfrac{{{e}^{x}}.{{x}^{2020}}}{{{e}^{x}}+1}\text{d}x}$
Suy ra, $I+I=\int\limits_{-2}^{2}{\dfrac{{{x}^{2020}}}{{{e}^{x}}+1}\text{d}x}+\int\limits_{-2}^{2}{\dfrac{{{e}^{x}}.{{x}^{2020}}}{{{e}^{x}}+1}\text{d}x}$
$\Leftrightarrow 2I=\int\limits_{-2}^{2}{\left[ \dfrac{{{x}^{2020}}}{{{e}^{x}}+1}\text{+}\dfrac{{{e}^{x}}.{{x}^{2020}}}{{{e}^{x}}+1} \right]\text{d}x}=\int\limits_{-2}^{2}{{{x}^{2020}}\text{d}x}=\left. \dfrac{{{x}^{2021}}}{2021} \right|_{-2}^{2}=\dfrac{{{2}^{2021}}}{2021}-\dfrac{{{\left( -2 \right)}^{2021}}}{2021}=\dfrac{{{2.2}^{2021}}}{2021}\Leftrightarrow I=\dfrac{{{2}^{2021}}}{2021}$.
Đổi cận
$x$ | $-2$ | $2$ |
$t$ | $2$ | $-2$ |
Suy ra, $I+I=\int\limits_{-2}^{2}{\dfrac{{{x}^{2020}}}{{{e}^{x}}+1}\text{d}x}+\int\limits_{-2}^{2}{\dfrac{{{e}^{x}}.{{x}^{2020}}}{{{e}^{x}}+1}\text{d}x}$
$\Leftrightarrow 2I=\int\limits_{-2}^{2}{\left[ \dfrac{{{x}^{2020}}}{{{e}^{x}}+1}\text{+}\dfrac{{{e}^{x}}.{{x}^{2020}}}{{{e}^{x}}+1} \right]\text{d}x}=\int\limits_{-2}^{2}{{{x}^{2020}}\text{d}x}=\left. \dfrac{{{x}^{2021}}}{2021} \right|_{-2}^{2}=\dfrac{{{2}^{2021}}}{2021}-\dfrac{{{\left( -2 \right)}^{2021}}}{2021}=\dfrac{{{2.2}^{2021}}}{2021}\Leftrightarrow I=\dfrac{{{2}^{2021}}}{2021}$.
Đáp án B.