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Step-by-Step Solution
Step 1: Use the Remainder Condition
We know that when a polynomial $P(x) = x^2 + bx + c$ is divided by $(x - 2)$, the remainder is simply $P(2)$. Given that this remainder equals 5, we get:
$P(2) = 2^2 + b \cdot 2 + c = 5.$
So,
$$
4 + 2b + c = 5 \quad\Longrightarrow\quad 2b + c = 1.
$$
Step 2: Set Up the Integral Condition
We are given:
$$
\int_{0}^{1} \bigl(x^2 + bx + c\bigr)\,dx = 1.
$$
Evaluating this definite integral:
$$
\int_{0}^{1} x^2\,dx = \left[\frac{x^3}{3}\right]_{0}^{1} = \frac{1}{3},
$$
$$
\int_{0}^{1} bx\,dx = b\left[\frac{x^2}{2}\right]_{0}^{1} = \frac{b}{2},
$$
$$
\int_{0}^{1} c\,dx = c[x]_{0}^{1} = c.
$$
Therefore,
$$
\frac{1}{3} + \frac{b}{2} + c = 1.
$$
Rearrange to get:
$$
\frac{b}{2} + c = 1 - \frac{1}{3} = \frac{2}{3}.
$$
Step 3: Solve the System of Equations
We now have two linear equations:
$2b + c = 1$
$\frac{b}{2} + c = \frac{2}{3}$
From the first equation, $c = 1 - 2b.$ Substitute this into the second:
$$
\frac{b}{2} + (1 - 2b) = \frac{2}{3}.
$$
Simplify:
$$
\frac{b}{2} + 1 - 2b = \frac{2}{3}
$$
$$
1 - \frac{3b}{2} = \frac{2}{3}.
$$
Move terms around:
$$
-\frac{3b}{2} = \frac{2}{3} - 1 = -\frac{1}{3}.
$$
Hence,
$$
\frac{3b}{2} = \frac{1}{3},
$$
$$
b = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}.
$$
Then
$$
c = 1 - 2\left(\frac{2}{9}\right) = 1 - \frac{4}{9} = \frac{5}{9}.
$$
Step 4: Compute the Required Expression
We are asked for $9(b + c)$. Substituting $b = \tfrac{2}{9}$ and $c = \tfrac{5}{9}$:
$$
b + c = \frac{2}{9} + \frac{5}{9} = \frac{7}{9},
$$
$$
9(b + c) = 9 \times \frac{7}{9} = 7.
$$
Therefore, the value of $9(b + c)$ is $\boxed{7}$.
