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Câu hỏi: Cho hàm số $y=f(x)$ liên tục trên $\mathbb{R}$ thỏa mãn $f(x)+x^{2019} f\left(x^{2020}\right)=\sqrt{1-x^2}$ với mọi $x$ thuộc $[0 ; 1]$. Tích phân $\int_0^1 f(x) d x$ bằng:
A. $1020604 \pi$.
B. $\dfrac{2017 \pi}{8072}$.
C. $\dfrac{505 \pi}{2021}$.
D. $\dfrac{\pi}{8076}$.
A. $1020604 \pi$.
B. $\dfrac{2017 \pi}{8072}$.
C. $\dfrac{505 \pi}{2021}$.
D. $\dfrac{\pi}{8076}$.
Có $\int_0^1 f(x) d x+\int_0^1 x^{2019} f\left(x^{2020}\right) d x=\int_0^1 \sqrt{1-x^2} d x=\dfrac{\pi}{4}$
Đặt $t=x^{2020} \Rightarrow d t=2020 . x^{2019} d x$ và $\int_0^1 x^{2019} f\left(x^{2020}\right) d x=\dfrac{1}{2020} \int_0^1 f(t) d t=\dfrac{1}{2020} \int_0^1 f(x) d x$
Vậy $\int_0^1 f(x) d x+\int_0^1 x^{2019} f\left(x^{2020}\right) d x=\dfrac{\pi}{4} \Leftrightarrow \dfrac{2021}{2020} \int_0^1 f(x) d x=\dfrac{\pi}{4} \Leftrightarrow \int_0^1 f(x) d x=\dfrac{505 \pi}{2021}$.
Đặt $t=x^{2020} \Rightarrow d t=2020 . x^{2019} d x$ và $\int_0^1 x^{2019} f\left(x^{2020}\right) d x=\dfrac{1}{2020} \int_0^1 f(t) d t=\dfrac{1}{2020} \int_0^1 f(x) d x$
Vậy $\int_0^1 f(x) d x+\int_0^1 x^{2019} f\left(x^{2020}\right) d x=\dfrac{\pi}{4} \Leftrightarrow \dfrac{2021}{2020} \int_0^1 f(x) d x=\dfrac{\pi}{4} \Leftrightarrow \int_0^1 f(x) d x=\dfrac{505 \pi}{2021}$.
Đáp án C.