Lời giải:
\(f'(x)=f'(x-1); dx=d(x-1)\)
\(\Rightarrow f'(x)dx=f'(x-1)d(x-1)\)
\(\Rightarrow \int f'(x)dx=\int f'(x-1)d(x-1)\)
\(\Rightarrow f(x)=f(x-1)+c\)
Thay $x=1$ ta có \(f(1)=f(0)+c\Leftrightarrow 2019=1+c\Rightarrow c=2018\)
Khi đó: $f(x)=f(x-1)+2018$
\(f(0)=1=1+2018.0\)
\(f(1)=1+2018.1\)
\(f(2)=f(1)+2018=1+2018.1+2018=1+2018.2\)
.........
\(\Rightarrow f(x)=1+2018.x\)
Do đó: \(\int ^{1}_{0}f(x)dx=\int ^{1}_{0}(2018x+1)dx=\left.\begin{matrix} 1\\ 0\end{matrix}\right|(1009x^2+x)=1010\)
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