Let `(d)/(dx)F (x)= (e^(sin x))/(x), x gt 0`. If `int_(1)^(4) (3x^2)/(x^3) e^(sin x^(3)) dx= F(k)- F(1)` then one possible value of k is

Options

A. 15
B. 16
C. 63
D. 64

Answer

To solve the problem step by step, we start from the given information and apply integration techniques.

### Step 1: Understand the given information
We have \( \frac{d}{dx} F(x) = \frac{e^{\sin x}}{x} \) for \( x > 0 \). We also have the integral:
\[ \int_{1}^{4} \frac{3x^2}{x^3} e^{\sin(x^3)} \, dx = F(k) - F(1) \]

### Step 2: Simplify the integral
First, simplify the integrand:
\[ \frac{3x^2}{x^3} = \frac{3}{x} \]
So, the integral becomes:
\[ \int_{1}^{4} \frac{3}{x} e^{\sin(x^3)} \, dx \]

### Step 3: Use substitution
Let \( t = x^3 \). Then, differentiate:
\[ dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2} \]
Now, we need to change the limits of integration. When \( x = 1 \), \( t = 1^3 = 1 \). When \( x = 4 \), \( t = 4^3 = 64 \).

### Step 4: Substitute in the integral
Substituting \( t \) into the integral:
\[ \int_{1}^{64} e^{\sin(t)} \frac{1}{t} \, dt \]
This integral can be recognized as:
\[ \int_{1}^{64} \frac{d}{dt} F(t) \, dt \]
where \( F(t) \) is an antiderivative of \( \frac{e^{\sin t}}{t} \).

### Step 5: Evaluate the integral using the Fundamental Theorem of Calculus
By the Fundamental Theorem of Calculus:
\[ \int_{1}^{64} \frac{d}{dt} F(t) \, dt = F(64) - F(1) \]

### Step 6: Set up the equation
From the problem statement, we have:
\[ F(64) - F(1) = F(k) - F(1) \]
This implies:
\[ F(64) = F(k) \]

### Step 7: Conclude the value of \( k \)
Since \( F \) is a function, if \( F(64) = F(k) \), then one possible value of \( k \) is:
\[ k = 64 \]

Thus, the final answer is:
\[ \boxed{64} \]