If $P\left(x\right)$ is a polynomial of the least degree that has a maximum equal to 6 at $x=1,$ and a minimum equal to 2 at $x=3,$ then ${\displaystyle {\int }_0^{1}P\left(x\right)dx}$ equals $\frac{19}{t},$  the value of ‘t’ is –––––––––––

$p^1\left(3\right)=0\to$  minimum

$p^1\left(x\right)=k\left(x-1\right)\left(x-3\right)$

$=k\left(x^2-4x+3\right)$

${\displaystyle \int p^1\left(dx\right)=}{\displaystyle \int k\left(x^2-4x+3\right)dx}$

$p\left(x\right)=k\left(\frac{x^3}{3}-4\frac{x^2}{2}+3x\right)+c$

$p\left(1\right)=6,p\left(3\right)=2$

$p\left(1\right)=k\left(\frac{1}{3}-2+3\right)+c=6$

$c+\frac{4k}{3}=6\to \left(1\right)$

$p\left(3\right)=2\Rightarrow k\left[\frac{27}{3}-18+9\right]+c=2$

$c=2\to \left(2\right)$

$\left(1\right)\&\left(2\right)\Rightarrow 2+\frac{4k}{3}=6\Rightarrow \frac{4k}{3}=4\Rightarrow k=3$

$p\left(x\right)=3\left(\frac{x^3}{3}-2x^2+3x\right)+2$

$=x^3-6x^2+9x+2$

${\displaystyle {\int }_0^{1}p\left(x\right)dx=}{\left(\frac{x^4}{4}-\frac{6x^3}{3}+\frac{9x^2}{2}+2x\right)}_0^1=\frac{1}{4}-\frac{6}{3}+\frac{9}{2}+2$

$=\frac{3-24+54+24}{12}$

$=\frac{57}{12}=\frac{19}{3}=\frac{19}{t}\Rightarrow t=3$