Let $$f : R \to R$$ be defined as $$f(x) = e^{-x}\sin x$$. If $$F : [0, 1] \to R$$ is a differentiable function such that $$F(x) = \int_0^x f(t)dt$$, then the value of $$\int_0^1 (F'(x) + f(x))e^x dx$$ lies in the interval:

We have $$f(x) = e^{-x}\sin x$$ and $$F(x) = \int_0^x f(t)\,dt$$. By the Fundamental Theorem of Calculus, $$F'(x) = f(x) = e^{-x}\sin x$$.

We need $$\int_0^1 (F'(x) + f(x))e^x\,dx$$. Since $$F'(x) = f(x)$$, this becomes $$\int_0^1 2f(x)e^x\,dx = \int_0^1 2e^{-x}\sin x \cdot e^x\,dx = \int_0^1 2\sin x\,dx$$.

Evaluating: $$\int_0^1 2\sin x\,dx = 2[-\cos x]_0^1 = 2(-\cos 1 + \cos 0) = 2(1 - \cos 1)$$.

Now $$\cos 1 \approx 0.5403$$, so $$2(1 - \cos 1) \approx 2(0.4597) = 0.9194$$.

Checking the intervals: $$\frac{330}{360} \approx 0.9167$$ and $$\frac{331}{360} \approx 0.9194$$. Since $$2(1 - \cos 1) \approx 0.9194$$, this value lies in the interval $$\left[\frac{330}{360}, \frac{331}{360}\right]$$.

The answer is $$\left[\frac{330}{360}, \frac{331}{360}\right]$$, which is Option B.