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Step 1: Identify the relationship between F(x) and f(x)
We are given that
$F(x) = \int_{0}^{x} f(t) \, dt.$
By the Fundamental Theorem of Calculus,
$F'(x) = f(x).$
Step 2: Express the integral to be evaluated
We need to compute
$\displaystyle I = \int_{0}^{1} \bigl(F'(x) + f(x)\bigr)e^x \, dx.$
Substituting $F'(x) = f(x)$ into the integrand, we get
\[
F'(x) + f(x) = f(x) + f(x) = 2\,f(x).
\]
Hence,
\[
I = \int_{0}^{1} 2\,f(x)\,e^x \, dx.
\]
Step 3: Simplify the integrand using f(x)
The function $f(x)$ is given as
$f(x) = e^{-x} \sin x.$
Therefore,
\[
2\,f(x)\,e^x = 2 \,\bigl(e^{-x}\sin x\bigr)\,e^x = 2\,\sin x.
\]
So the integral simplifies to
\[
I = \int_{0}^{1} 2 \,\sin x \, dx.
\]
Step 4: Evaluate the simplified integral
We know that
\[
\int \sin x \, dx = -\cos x.
\]
Thus,
\[
I = \int_{0}^{1} 2\sin x \, dx
= 2 \bigl[-\cos x\bigr]_{0}^{1}
= 2 \bigl[-\cos(1) + \cos(0)\bigr]
= 2 \bigl[1 - \cos(1)\bigr].
\]
Step 5: Determine the interval for the value of the integral
The problem states that $2(1 - \cos 1)$ lies in the interval
$\bigl[\tfrac{330}{360},\, \tfrac{331}{360}\bigr].$ Numerically,
\[
\cos(1) \approx 0.5403 \quad (\text{where 1 is in radians}),
\]
which makes
\[
1 - \cos(1) \approx 1 - 0.5403 = 0.4597,
\]
and hence
\[
2(1 - \cos(1)) \approx 2 \times 0.4597 = 0.9194.
\]
We note
\[
330/360 \approx 0.9167 \quad \text{and} \quad 331/360 \approx 0.9194,
\]
so indeed
\[
2(1 - \cos(1)) \in \bigl[0.9167,\, 0.9194\bigr],
\]
which corresponds to
$\displaystyle \left[\frac{330}{360}, \frac{331}{360}\right].$
