Let $P\left(x\right)=x^2+bx+c$ be a quadratic polynomial with real coefficients such that $\underset{0}{\overset{1}{\int }}P\left(x\right)dx=1$ and $P\left(x\right)$  leaves remainder 5 when it is divided by $\left(x-2\right).$ Then the value of  is $9\left(b+c\right)$ equal to

$\begin{array}{l}{\int }_0^1\left(x^2+bx+c\right)dx=1\\ \frac{x^3}{3}+\frac{b{x}^2}{2}+{\left.cx\right|}_0^1=1\\ \frac{1}{3}+\frac{b}{2}+c=1\\ 3b+6c=5 ...... \left(1\right)\\ Given the remianinder when P\left(x\right) is divided by (x-2) is 5\\ It means P\left(2\right) =5\\ 4+2b+c=5\frac{8}{3}+\frac{4b}{2}+2c=5\\ 2b+c=1 ......\left(2\right)\\ \\ \text{ Solving } b=\frac{2}{9},c=\frac{5}{9},b+c=\frac{7}{9},9(b+c)=7\\ \end{array}$