Key Concepts and Formulas

• Fundamental Theorem of Calculus (Part 1): If $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$. This theorem allows us to find the derivative of an integral with a variable upper limit.
• Integration by Parts: The formula for integration by parts is $\int u \, dv = uv - \int v \, du$. This is useful for integrating products of functions.
• Properties of Exponential and Trigonometric Functions: Understanding the behavior and series expansions of $e^{-x}$ and $\sin x$ can be helpful in evaluating integrals or series.

Step-by-Step Solution

Step 1: Analyze the given information and the integral to be evaluated. We are given a function $f(x) = e^{-x} \sin x$ and a differentiable function $F(x) = \int_0^x f(t) dt$. We need to find the value of the integral $\int_0^1 (F'(x) + f(x))e^x dx$.

Step 2: Apply the Fundamental Theorem of Calculus to find $F'(x)$. According to the Fundamental Theorem of Calculus (Part 1), if $F(x) = \int_0^x f(t) dt$, then its derivative $F'(x)$ is equal to the integrand evaluated at the upper limit, i.e., $F'(x) = f(x)$. Substituting the given $f(x)$, we get $F'(x) = e^{-x} \sin x$.

Step 3: Substitute $F'(x)$ into the integral. The integral we need to evaluate is $\int_0^1 (F'(x) + f(x))e^x dx$. Substituting $F'(x) = f(x) = e^{-x} \sin x$, the integral becomes: $\int_0^1 (e^{-x} \sin x + e^{-x} \sin x)e^x dx$ $\int_0^1 (2e^{-x} \sin x)e^x dx$

Step 4: Simplify the integrand. When we multiply $e^{-x}$ by $e^x$, they cancel each other out: $e^{-x} e^x = e^{-x+x} = e^0 = 1$. So, the integrand simplifies to $2 \sin x$. The integral now is: $\int_0^1 2 \sin x dx$

Step 5: Evaluate the definite integral. We need to find the antiderivative of $2 \sin x$. The antiderivative of $\sin x$ is $-\cos x$. So, the antiderivative of $2 \sin x$ is $-2 \cos x$. Now, we evaluate this antiderivative from $0$ to $1$: $[-2 \cos x]_0^1 = (-2 \cos 1) - (-2 \cos 0)$ $= -2 \cos 1 + 2 \cos 0$ Since $\cos 0 = 1$, we have: $= -2 \cos 1 + 2(1)$ $= 2 - 2 \cos 1$

Step 6: Approximate the value of the integral. We need to find the numerical value of $2 - 2 \cos 1$. We know that $1$ radian is approximately $57.3^\circ$. We can use the Taylor series expansion for $\cos x$ around $x=0$: $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$ For $x=1$ radian: $\cos 1 \approx 1 - \frac{1^2}{2!} + \frac{1^4}{4!} - \frac{1^6}{6!}$ $\cos 1 \approx 1 - \frac{1}{2} + \frac{1}{24} - \frac{1}{720}$ $\cos 1 \approx 1 - 0.5 + 0.041666... - 0.001388...$ $\cos 1 \approx 0.540277...$ Now, substitute this back into the expression for the integral: $2 - 2 \cos 1 \approx 2 - 2(0.540277...)$ $\approx 2 - 1.080554...$ $\approx 0.919446...$

Step 7: Convert the decimal approximation to a fraction and compare with the options. We need to compare $0.919446...$ with the given fractional options. Let's convert the options to decimals. The common denominator is 360. (A) $\frac{331}{360} \approx 0.919444...$ (B) $\frac{330}{360} \approx 0.916666...$ (C) $\frac{335}{360} \approx 0.930555...$ (D) $\frac{327}{360} \approx 0.908333...$

Our calculated value $0.919446...$ is very close to $\frac{331}{360}$.

Alternative approach for Step 4 & 5 (using integration by parts on the original integral): Consider the integral $I = \int_0^1 (F'(x) + f(x))e^x dx$. We know $F'(x) = f(x)$, so $I = \int_0^1 2 f(x) e^x dx = \int_0^1 2 e^{-x} \sin x e^x dx = \int_0^1 2 \sin x dx$. This leads to the same result as above.

Let's consider a slightly different manipulation of the integral: $I = \int_0^1 F'(x)e^x dx + \int_0^1 f(x)e^x dx$

For the first part, $\int_0^1 F'(x)e^x dx$, we can use integration by parts with $u = e^x$ and $dv = F'(x)dx$. Then $du = e^x dx$ and $v = F(x)$. $\int_0^1 F'(x)e^x dx = [e^x F(x)]_0^1 - \int_0^1 F(x)e^x dx$ $= (e^1 F(1) - e^0 F(0)) - \int_0^1 F(x)e^x dx$ Since $F(0) = \int_0^0 f(t) dt = 0$, this becomes: $= e F(1) - \int_0^1 F(x)e^x dx$

For the second part, $\int_0^1 f(x)e^x dx$: We know $f(x) = e^{-x} \sin x$, so $f(x)e^x = e^{-x} \sin x e^x = \sin x$. Thus, $\int_0^1 f(x)e^x dx = \int_0^1 \sin x dx = [-\cos x]_0^1 = -\cos 1 - (-\cos 0) = 1 - \cos 1$.

Now, let's look at the definition of $F(1)$: $F(1) = \int_0^1 f(t) dt = \int_0^1 e^{-t} \sin t dt$. The integral $\int e^{-x} \sin x dx$ can be solved using integration by parts twice. Let $J = \int e^{-x} \sin x dx$. Let $u = \sin x, dv = e^{-x} dx$. Then $du = \cos x dx, v = -e^{-x}$. $J = -e^{-x} \sin x - \int (-e^{-x}) \cos x dx = -e^{-x} \sin x + \int e^{-x} \cos x dx$. Now, let $u = \cos x, dv = e^{-x} dx$. Then $du = -\sin x dx, v = -e^{-x}$. $J = -e^{-x} \sin x + (-e^{-x} \cos x - \int (-e^{-x}) (-\sin x) dx)$ $J = -e^{-x} \sin x - e^{-x} \cos x - \int e^{-x} \sin x dx$ $J = -e^{-x} (\sin x + \cos x) - J$ $2J = -e^{-x} (\sin x + \cos x)$ $J = -\frac{1}{2} e^{-x} (\sin x + \cos x)$

So, $F(1) = \int_0^1 e^{-t} \sin t dt = \left[-\frac{1}{2} e^{-t} (\sin t + \cos t)\right]_0^1$ $F(1) = \left(-\frac{1}{2} e^{-1} (\sin 1 + \cos 1)\right) - \left(-\frac{1}{2} e^0 (\sin 0 + \cos 0)\right)$ $F(1) = -\frac{1}{2e} (\sin 1 + \cos 1) + \frac{1}{2} (0 + 1)$ $F(1) = \frac{1}{2} - \frac{1}{2e} (\sin 1 + \cos 1)$

The original integral is $I = \int_0^1 (F'(x) + f(x))e^x dx = \int_0^1 (e^{-x} \sin x + e^{-x} \sin x) e^x dx = \int_0^1 2 \sin x dx = 2 - 2 \cos 1$. This confirms our earlier simplified approach is correct and much more direct.

Let's re-evaluate the approximation for $2 - 2 \cos 1$ with more precision. Using a calculator, $\cos(1 \text{ radian}) \approx 0.540302305868$. $2 - 2 \cos 1 \approx 2 - 2 \times 0.540302305868$ $2 - 2 \cos 1 \approx 2 - 1.080604611736$ $2 - 2 \cos 1 \approx 0.919395388264$

Now, let's check the fractions again: (A) $\frac{331}{360} \approx 0.919444444444$ (B) $\frac{330}{360} \approx 0.916666666666$ (C) $\frac{335}{360} \approx 0.930555555555$ (D) $\frac{327}{360} \approx 0.908333333333$

Our calculated value $0.919395...$ is indeed closest to $\frac{331}{360}$.

Common Mistakes & Tips

• Misapplying the Fundamental Theorem of Calculus: Ensure you correctly identify $F'(x)$ as $f(x)$ when $F(x) = \int_a^x f(t) dt$.
• Algebraic Errors with Exponentials: Be careful when multiplying exponential terms like $e^{-x}$ and $e^x$. Remember $e^a e^b = e^{a+b}$.
• Approximation Accuracy: When dealing with options that are close fractions, use a sufficiently accurate value for trigonometric functions or perform the calculations to maintain precision.

Summary

The problem asks us to evaluate the definite integral $\int_0^1 (F'(x) + f(x))e^x dx$, where $F(x) = \int_0^x f(t) dt$ and $f(x) = e^{-x} \sin x$. By applying the Fundamental Theorem of Calculus, we found that $F'(x) = f(x)$. Substituting this into the integral, the expression simplifies significantly due to the cancellation of exponential terms, leading to $\int_0^1 2 \sin x dx$. Evaluating this integral gives $2 - 2 \cos 1$. Approximating this value and comparing it with the given fractional options, we found it to be closest to $\frac{331}{360}$.

Final Answer

The final answer is \boxed{\left[ {{331} \over {360}},{{334} \over {360}} \right]}.