Question Number 224085 by MathematicalUser2357 last updated on 18/Aug/25
$$\mathrm{If}\:{f}\left({x}\right)=\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +{x},\:\mathrm{Then}\:\mathrm{solve}\:\mathrm{for}\:{a}\:\mathrm{and}\:{b}: \\ $$$$\underset{{x}\in\mathbb{R}} {\mathrm{max}}\left\{\int_{{x}} ^{\mathrm{2}} {f}\left({t}\right){dt}\right\}={a}\:\mathrm{where}\:{x}={b} \\ $$
Answered by Ghisom_ last updated on 18/Aug/25
![(d/dx)[∫_x ^2 f(t)dt]=0∧(d^2 /dx^2 )[∫_x ^2 f(t)dt]<0 (d/dx)[∫_x ^2 f(t)dt]=−lim_(t→x) f(t) =−(4x^3 +3x^2 +x)=0 ⇒ x=0 ⇒ (d^2 /dx^2 )[∫_x ^2 f(t)dt]=−12x^2 −6x−1<0 ⇒ max (∫_x ^2 f(t)dt)=26 (a; b)=(26; 0)](https://www.tinkutara.com/question/Q224090.png)
$$\frac{{d}}{{dx}}\left[\underset{{x}} {\overset{\mathrm{2}} {\int}}{f}\left({t}\right){dt}\right]=\mathrm{0}\wedge\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\left[\underset{{x}} {\overset{\mathrm{2}} {\int}}{f}\left({t}\right){dt}\right]<\mathrm{0} \\ $$$$\frac{{d}}{{dx}}\left[\underset{{x}} {\overset{\mathrm{2}} {\int}}{f}\left({t}\right){dt}\right]=−\underset{{t}\rightarrow{x}} {\mathrm{lim}}\:{f}\left({t}\right)\:=−\left(\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +{x}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\mathrm{0} \\ $$$$\Rightarrow\:\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\left[\underset{{x}} {\overset{\mathrm{2}} {\int}}{f}\left({t}\right){dt}\right]=−\mathrm{12}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{1}<\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{max}\:\left(\underset{{x}} {\overset{\mathrm{2}} {\int}}{f}\left({t}\right){dt}\right)=\mathrm{26} \\ $$$$\left({a};\:{b}\right)=\left(\mathrm{26};\:\mathrm{0}\right) \\ $$
Commented by MathematicalUser2357 last updated on 19/Aug/25
$${are}\:{you}\:{okay} \\ $$
Commented by RedstoneGG4 last updated on 19/Aug/25
$$\mathrm{Lmao} \\ $$
Answered by RedstoneGG4 last updated on 21/Aug/25
$$\int{f}\left({x}\right){dx}\:=\:{x}^{\mathrm{4}} \:+\:{x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:+\:{C} \\ $$$$\int_{{x}} ^{\mathrm{2}} {f}\left({t}\right){dt}\:=\:\mathrm{2}^{\mathrm{4}} \:+\:\mathrm{2}^{\mathrm{3}} \:+\:\cancel{\frac{\mathrm{1}}{\mathrm{2}}\centerdot}\mathrm{2}^{\cancel{\mathrm{2}}} \:−\:{x}^{\mathrm{4}} \:−\:{x}^{\mathrm{3}} \:−\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:=\:-{x}^{\mathrm{4}} \:−\:{x}^{\mathrm{3}} \:−\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:+\:\mathrm{26}\:=\:{g}\left({x}\right) \\ $$$$\underset{{x}\rightarrow\pm\infty} {\mathrm{lim}}\left(-{x}^{\mathrm{4}} \:\cancel{−\:{x}^{\mathrm{3}} \:−\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \:+\:\mathrm{26}}\right)\:=\:-\infty \\ $$$$\frac{{d}}{{dx}}\:{g}\left({x}\right)\:=\:\mathrm{0}\:\rightarrow\:-\mathrm{4}{x}^{\mathrm{3}} \:−\:\mathrm{3}{x}^{\mathrm{2}} \:−\:{x}\:=\:\mathrm{0}\:\rightarrow\:{x}\left(\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\mathrm{1}\:\neq\:\mathrm{0}\:\Leftarrow\:{x}\:\in\:\mathbb{R} \\ $$$${x}\:\in\:\mathbb{R}:\:\frac{{d}}{{dx}}\:{g}\left({x}\right)\:=\:\mathrm{0}\:\Leftrightarrow\:{x}\:=\:\mathrm{0} \\ $$$${x}\:=\:\mathrm{0}:\:\frac{{d}^{\mathrm{2}} }{{dx}^{\mathrm{2}} }\:{g}\left({x}\right)\:=\:-\mathrm{12}{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:−\:\mathrm{1}\:=\:-\mathrm{1}\:<\:\mathrm{0} \\ $$$${b}\:=\:\mathrm{0} \\ $$$${g}\left({b}\right)\:=\:\cancel{-\mathrm{0}^{\mathrm{4}} \:−\:\mathrm{0}^{\mathrm{3}} \:−\:\frac{\mathrm{1}}{\mathrm{2}}\centerdot\mathrm{0}^{\mathrm{2}} \:+}\:\mathrm{26}\:=\:\mathrm{26} \\ $$$${a}\:=\:\mathrm{26} \\ $$