Suppose that F (x) is an antiderivative of `f (x)=sinx/x,x>0` , then `int_1^3 (sin2x)/x dx` can be expressed as

Options

A. `F(6)-F(2)`
B. `1/2(F(6)-F(2))`
C. `1/2(F(3)-F(1))`
D. `2(F(6)-F(2))`

Explanation

To solve the problem, we need to evaluate the integral

\[ I = \int_1^3 \frac{\sin(2x)}{x} \, dx \]

and express it in terms of the antiderivative \( F(x) \) of \( f(x) = \frac{\sin x}{x} \).

### Step 1: Use Substitution
Let's use the substitution \( t = 2x \). Then, we differentiate to find \( dx \):

\[ dt = 2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{2} \]

### Step 2: Change the Limits of Integration
When \( x = 1 \):

\[ t = 2 \cdot 1 = 2 \]

When \( x = 3 \):

\[ t = 2 \cdot 3 = 6 \]

### Step 3: Substitute in the Integral
Now we can rewrite the integral \( I \):

\[ I = \int_1^3 \frac{\sin(2x)}{x} \, dx = \int_2^6 \frac{\sin(t)}{t/2} \cdot \frac{dt}{2} \]

This simplifies to:

\[ I = \int_2^6 \frac{\sin(t)}{t} \, dt \]

### Step 4: Express in Terms of Antiderivative
Since \( F(x) \) is the antiderivative of \( f(x) = \frac{\sin x}{x} \), we can express the integral as:

\[ I = F(6) - F(2) \]

### Final Result
Thus, the integral \( \int_1^3 \frac{\sin(2x)}{x} \, dx \) can be expressed as:

\[ I = F(6) - F(2) \]