Question

Let the area bounded by the curve $y=a^2{x}^2+ax+1(a\ne 0)$ and the straight lines $y=0,x=0$ and $x=1$ is least, then the absolute value of $4a$ is

Answer 1

Given equation is $y=a^2{x}^2+ax+1$
Coefficient of $x^2>0$ and $D=a^2-4a^2<0$
$\therefore$ The graph of the given curve and the straight lines is :


Let $A$ be the area bounded :
$A=\underset{0}{\overset{1}{\int }}(a^2{x}^2+ax+1)dx$
$A={\frac{a^2{x}^3}{3}+\frac{a{x}^2}{2}+x|}_0^1$
$A=\frac{a^2}{3}+\frac{a}{2}+1$
For $A_{min}$
$\frac{dA}{da}=0$
$\Rightarrow \frac{2a}{3}+\frac{1}{2}=0$
$\therefore a=\frac{-3}{4}$