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Câu hỏi: Cho hàm số $f\left( x \right)={{\left| x \right|}^{2021}}$. Giá trị của $I=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( 2\cos x-1 \right)\sin x\text{d}x}$ bằng:
A. $I=\dfrac{1}{2022}$.
B. $I=\dfrac{1}{2021}$.
C. $I=\dfrac{1}{4042}$.
D. $I=0$.
A. $I=\dfrac{1}{2022}$.
B. $I=\dfrac{1}{2021}$.
C. $I=\dfrac{1}{4042}$.
D. $I=0$.
+ Đặt $t=2\cos x-1\Rightarrow \text{dt}=-2\sin x\text{d}x\Rightarrow \sin x\text{d}x=-\dfrac{\text{dt}}{2}$.
+ Khi $x=0$ thì $t=1$
Khi $x=\dfrac{\pi }{2}$ thì $t=-1$
+ Do đó: $I=\dfrac{1}{2}\int\limits_{-1}^{1}{f(t)\text{dt}=}\dfrac{1}{2}\int\limits_{-1}^{1}{f(x)\text{d}x=\dfrac{1}{2}\int\limits_{-1}^{1}{{{\left| x \right|}^{2021}}\text{d}x}}$.
Vì $f(x)={{\left| x \right|}^{2021}}$ là hàm số chẵn nên $\int\limits_{-1}^{1}{{{\left| x \right|}^{2021}}dx=}2\int\limits_{0}^{1}{{{\left| x \right|}^{2021}}dx}$.
Suy ra $I=\int\limits_{0}^{1}{{{\left| x \right|}^{2021}}dx=\int\limits_{0}^{1}{{{x}^{2021}}dx=}}\left. \dfrac{{{x}^{2022}}}{2022} \right|_{0}^{1}=\dfrac{1}{2022}$.
+ Khi $x=0$ thì $t=1$
Khi $x=\dfrac{\pi }{2}$ thì $t=-1$
+ Do đó: $I=\dfrac{1}{2}\int\limits_{-1}^{1}{f(t)\text{dt}=}\dfrac{1}{2}\int\limits_{-1}^{1}{f(x)\text{d}x=\dfrac{1}{2}\int\limits_{-1}^{1}{{{\left| x \right|}^{2021}}\text{d}x}}$.
Vì $f(x)={{\left| x \right|}^{2021}}$ là hàm số chẵn nên $\int\limits_{-1}^{1}{{{\left| x \right|}^{2021}}dx=}2\int\limits_{0}^{1}{{{\left| x \right|}^{2021}}dx}$.
Suy ra $I=\int\limits_{0}^{1}{{{\left| x \right|}^{2021}}dx=\int\limits_{0}^{1}{{{x}^{2021}}dx=}}\left. \dfrac{{{x}^{2022}}}{2022} \right|_{0}^{1}=\dfrac{1}{2022}$.
Đáp án A.