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Question Number 201463 by hardmath last updated on 06/Dec/23
a = constant number:  if   ∫x ∙ f(x) dx = x^3  - x^2  + 4x - (a/5)  find   f(2) = ?
$$\mathrm{a}\:=\:\mathrm{constant}\:\mathrm{number}: \\ $$$$\mathrm{if}\:\:\:\int\mathrm{x}\:\centerdot\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:\mathrm{x}^{\mathrm{3}} \:-\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:-\:\frac{\mathrm{a}}{\mathrm{5}} \\ $$$$\mathrm{find}\:\:\:\mathrm{f}\left(\mathrm{2}\right)\:=\:? \\ $$
Answered by mr W last updated on 06/Dec/23
definition:  ∫g(x)dx=G(x) means  g(x)=((dG(x))/dx)    xf(x)=(d/dx)(x^3 −x^2 +4x−(a/5))  xf(x)=3x^2 −2x+4  ⇒f(x)=3x−2+(4/x)  ⇒f(2)=3×2−2+(4/2)=6 ✓
$${definition}: \\ $$$$\int{g}\left({x}\right){dx}={G}\left({x}\right)\:{means} \\ $$$${g}\left({x}\right)=\frac{{dG}\left({x}\right)}{{dx}} \\ $$$$ \\ $$$${xf}\left({x}\right)=\frac{{d}}{{dx}}\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{4}{x}−\frac{{a}}{\mathrm{5}}\right) \\ $$$${xf}\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{3}{x}−\mathrm{2}+\frac{\mathrm{4}}{{x}} \\ $$$$\Rightarrow{f}\left(\mathrm{2}\right)=\mathrm{3}×\mathrm{2}−\mathrm{2}+\frac{\mathrm{4}}{\mathrm{2}}=\mathrm{6}\:\checkmark \\ $$
Commented by hardmath last updated on 08/Dec/23
cool dear professor thank you
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$

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