Let d/(dx)F(x)=((e^(sinx))/x),x > 0. If int_1^4 3/x e^sin (x^3)dx=F(k)-F(1), then one of the possible values of k , is: (a) 15 (b) 16 (c) 63 (d) 64

Options

A. 27
B. 18
C. 9
D. none of these

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the given information
We have the derivative of a function \( F(x) \) given by:
\[ \frac{d}{dx} F(x) = \frac{e^{\sin x}}{x}, \quad x > 0 \]
We also have the integral:
\[ \int_1^4 \frac{3}{x} e^{\sin (x^3)} \, dx = F(k) - F(1) \]

### Step 2: Rewrite the integral
We can manipulate the integral by multiplying and dividing by \( x^2 \):
\[ \int_1^4 \frac{3}{x} e^{\sin (x^3)} \, dx = \int_1^4 \frac{3x^2}{x^3} e^{\sin (x^3)} \, dx \]

### Step 3: Use substitution
Let \( t = x^3 \). Then, differentiating gives us:
\[ dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2} \]
Also, when \( x = 1 \), \( t = 1^3 = 1 \) and when \( x = 4 \), \( t = 4^3 = 64 \).

### Step 4: Change the limits and the integral
Substituting in the integral gives:
\[ \int_1^{64} \frac{e^{\sin t}}{t} \, dt \]
Thus, we can rewrite the original integral as:
\[ \int_1^{64} \frac{e^{\sin t}}{t} \, dt = F(k) - F(1) \]

### Step 5: Relate the integrals to \( F(k) \)
From the Fundamental Theorem of Calculus, we know:
\[ \int_1^{64} \frac{d}{dt} F(t) \, dt = F(64) - F(1) \]
This means:
\[ F(64) - F(1) = F(k) - F(1) \]

### Step 6: Solve for \( k \)
From the equation \( F(64) - F(1) = F(k) - F(1) \), we can deduce that:
\[ F(k) = F(64) \]
This implies:
\[ k = 64 \]

### Conclusion
Thus, one of the possible values of \( k \) is:
\[ \boxed{64} \]