Cho hàm số $f(x)$ có $f(1)=0$ và $f^{\prime }(x)=2019\cdot 2020\cdot x(x-1{)}^{2018},\forall x\in \mathbb{R}$. Khi đó ${\int }_0^{1}f(x)dx$ bằng

Chọn C.
Áp dụng phương pháp tích phân từng phần:
Đặt $\{\begin{array}{l}u=f(x)\\ dv=dx\end{array}⟹\{\begin{array}{l}du=f^{\prime }(x)dx\\ v=x\end{array}$.
Ta có:
${\int }_0^{1}f(x)dx={xf(x)|}_0^1-{\int }_0^{1}x{f}^{\prime }(x)dx=1\cdot f(1)-0\cdot f(0)-{\int }_0^{1}x{f}^{\prime }(x)dx=-{\int }_0^{1}x{f}^{\prime }(x)dx$ (do $f(1)=0$).
Thay $f^{\prime }(x)=2019\cdot 2020\cdot x(x-1{)}^{2018}$, ta được:
${\int }_0^{1}f(x)dx=-2019\cdot 2020{\int }_0^1{x}^2(x-1{)}^{2018}dx$.
Đặt $t=1-x⟹x=1-t,dx=-dt$. Khi $x=0⟹t=1$; $x=1⟹t=0$.
Ta có: $(x-1{)}^{2018}=(-t{)}^{2018}=t^{2018}$.
Khi đó:
${\int }_0^1{x}^2(x-1{)}^{2018}dx={\int }_0^1(1-t{)}^2{t}^{2018}dt={\int }_0^1(1-2t+t^2)t^{2018}dt$
$={\int }_0^1(t^{2018}-2t^{2019}+t^{2020})dt=\frac{1}{2019}-\frac{2}{2020}+\frac{1}{2021}$.
Ta tính:
$\frac{1}{2019}-\frac{1}{2020}=\frac{1}{2019\cdot 2020}$,
$-\frac{1}{2020}+\frac{1}{2021}=-\frac{1}{2020\cdot 2021}$.
Do đó tích phân bằng $\frac{1}{2019\cdot 2020}-\frac{1}{2020\cdot 2021}=\frac{2021-2019}{2019\cdot 2020\cdot 2021}=\frac{2}{2019\cdot 2020\cdot 2021}$.
Nhân với $-2019\cdot 2020$:
${\int }_0^{1}f(x)dx=-2019\cdot 2020\cdot \frac{2}{2019\cdot 2020\cdot 2021}=-\frac{2}{2021}$.