Question

If the area $A$ bounded by the curve $y=x^2+2x-3$ and the line $y=kx+1$ is minimum, then the value of $6A$ is

Answer 1

$x^2+2x-3=kx+1$
$\Rightarrow x^2+(2-k)x-4=0$
Let $x_1$ and $x_2$ be the roots of above equation.
$\Rightarrow x_1+x_2=k-2,x_1{x}_2=-4$
Then $A=\underset{x_1}{\overset{x_2}{\int }}\left[\right(kx+1)-(x^2+2x-3\left)\right]dx$
$={[4x+(k-2)\frac{x^2}{2}-\frac{x^3}{3}]}_{x_1}^{x_2}$
$=4(x_2-x_1)+\frac{k-2}{2}(x_2^2-x_1^2)-\frac{1}{3}(x_2^3-x_1^3)$
$=(x_2-x_1)[4+\frac{k-2}{2}(x_2+x_1)-\frac{1}{3}(x_2^2+x_1^2+x_1{x}_2\left)\right]$
$=\sqrt{(x_2+x_1{)}^2-4x_1{x}_2}[4+\frac{k-2}{2}(x_2+x_1)-\frac{1}{3}\left(\right(x_2+x_1{)}^2-x_1{x}_2)]$
$=\sqrt{(k-2{)}^2+16}[4+\frac{(k-2{)}^2}{2}-\frac{1}{3}\left(\right(k-2{)}^2+4)]$
$=\sqrt{(k-2{)}^2+16}[\frac{(k-2{)}^2}{6}+\frac{8}{3}]$

$\therefore A$ is minimum when $k=2$
and $A_{min}=\sqrt{16}\times \frac{8}{3}=\frac{32}{3}$