`f(x)=e^-xsinx , F(x)=int_0^xf(t)dt ,`.Find `int_0^1e^x(F'(x)+f(x))dx ` lies in interval

Options

A. `(330/360,331/360)`
B. `(327/360,329/360)`
C. `(335/360,336/360)`
D. None of these

Answer

To solve the problem, we need to evaluate the integral

\[ I = \int_0^1 e^x (F'(x) + f(x)) \, dx \]

where \( f(x) = e^{-x} \sin x \) and \( F(x) = \int_0^x f(t) \, dt \).

### Step 1: Find \( F'(x) \)

By the Fundamental Theorem of Calculus, we know that:

\[ F'(x) = f(x) \]

Thus, we have:

\[ F'(x) = e^{-x} \sin x \]

### Step 2: Substitute \( F'(x) \) into the integral

Now we can substitute \( F'(x) \) into the integral \( I \):

\[ I = \int_0^1 e^x (f(x) + f(x)) \, dx = \int_0^1 e^x (2f(x)) \, dx \]

This simplifies to:

\[ I = 2 \int_0^1 e^x f(x) \, dx \]

### Step 3: Substitute \( f(x) \)

Now substitute \( f(x) \):

\[ I = 2 \int_0^1 e^x (e^{-x} \sin x) \, dx = 2 \int_0^1 \sin x \, dx \]

### Step 4: Evaluate the integral

Now we need to evaluate the integral \( \int_0^1 \sin x \, dx \):

\[ \int \sin x \, dx = -\cos x + C \]

Thus,

\[ \int_0^1 \sin x \, dx = [-\cos x]_0^1 = -\cos(1) - (-\cos(0)) = -\cos(1) + 1 \]

So,

\[ \int_0^1 \sin x \, dx = 1 - \cos(1) \]

### Step 5: Substitute back into \( I \)

Substituting this back into our expression for \( I \):

\[ I = 2(1 - \cos(1)) = 2 - 2\cos(1) \]

### Step 6: Determine the interval

Now we need to find the interval in which \( I \) lies. We know that \( \cos(1) \) is a positive number less than 1. Therefore, \( 2\cos(1) \) is a positive number less than 2, and thus:

\[ 0 < 2 - 2\cos(1) < 2 \]

To find a more precise interval, we can use the approximation \( \cos(1) \approx 0.5403 \):

\[ I \approx 2 - 2(0.5403) = 2 - 1.0806 = 0.9194 \]

We can also calculate the bounds:

- When \( \cos(1) = 1 \), \( I = 0 \)
- When \( \cos(1) = 0 \), \( I = 2 \)

Thus, the value of \( I \) lies between \( 0 \) and \( 2 \).

### Final Result

The integral \( I \) lies in the interval \( (0, 2) \).