Let $$P(x) = x^2 + bx + c$$ be a quadratic polynomial with real coefficients such that $$\int_0^1 P(x)dx = 1$$ and $$P(x)$$ leaves remainder 5 when it is divided by $$(x-2)$$. Then the value of $$9(b+c)$$ is equal to:

We have $$P(x) = x^2 + bx + c$$ with two conditions: $$\int_0^1 P(x) \, dx = 1$$ and $$P(x)$$ leaves remainder 5 when divided by $$(x-2)$$, meaning $$P(2) = 5$$.

From the integral condition: $$\int_0^1 (x^2 + bx + c) \, dx = \frac{1}{3} + \frac{b}{2} + c = 1$$, so $$\frac{b}{2} + c = \frac{2}{3}$$, which gives $$3b + 6c = 4$$ ... (i).

From the remainder condition: $$P(2) = 4 + 2b + c = 5$$, so $$2b + c = 1$$ ... (ii).

From (ii): $$c = 1 - 2b$$. Substituting into (i): $$3b + 6(1 - 2b) = 4$$, so $$3b + 6 - 12b = 4$$, giving $$-9b = -2$$, hence $$b = \frac{2}{9}$$.

Then $$c = 1 - \frac{4}{9} = \frac{5}{9}$$.

Therefore, $$9(b + c) = 9\left(\frac{2}{9} + \frac{5}{9}\right) = 9 \cdot \frac{7}{9} = 7$$.