Answer
$$4x-\frac{1}{2}x^2+C$$
Work Step by Step
$\int 4-x$ $dx$
$=4\frac{1}{0+1}x^{0+1}\frac{1}{1+1}x^{1+1}+C$
$=4x-\frac{1}{2}x^2+C$
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