Question: Let $\frac{d}{dx}(f(x))=e^{sinx}, x>0$ if $\int_{1}^{4} 2 \frac{e^{sinx^2}}{x}dx=$ $F(K)-f(1)$ then ...

Question

Let $\frac{d}{dx}(f(x))=e^{sinx}, x>0$ if $\int_{1}^{4} 2 \frac{e^{sinx^2}}{x}dx=$ $F(K)-f(1)$ then one of the possible value of K is

Answer 1

Solution

The problem asks us to find a possible value of K given the derivative of a function f(x) and an equality involving a definite integral.

The given information is:

d/dx(f(x)) = e^(sin x), for x > 0.
∫_{1}^{4} 2 \frac{e^{sinx^2}}{x}dx = F(K)-f(1).

There appears to be a slight inconsistency in the problem statement when compared to standard calculus problems of this type, especially in light of the provided "similar question". If d/dx(f(x)) = e^(sin x), then f(x) is an antiderivative of e^(sin x). Let's evaluate the given integral I = ∫_{1}^{4} 2 \frac{e^{sinx^2}}{x}dx.

To simplify this integral, we use the substitution method. Let u = x^2. Then, differentiate u with respect to x: du/dx = 2x. So, du = 2x dx.

Now, we need to rewrite the integrand in terms of u and du. The integrand is 2 * e^(sin(x^2)) / x. We can rewrite 2/x as (2x) / x^2. So, the integrand becomes (2x / x^2) * e^(sin(x^2)). This can be further written as (1 / x^2) * e^(sin(x^2)) * (2x dx).

Now, substitute u = x^2 and du = 2x dx: The term x^2 in the denominator becomes u. The term e^(sin(x^2)) becomes e^(sin u). The term 2x dx becomes du.

Next, change the limits of integration according to the substitution: When x = 1, u = 1^2 = 1. When x = 4, u = 4^2 = 16.

So, the integral I transforms to: I = ∫_{1}^{16} \frac{1}{u} e^{sin u} du

Now, let's look at the given d/dx(f(x)) = e^(sin x). If this is strictly followed, then f'(u) = e^(sin u). However, the integral we obtained is ∫_{1}^{16} \frac{e^{sin u}}{u} du. This integral's integrand is e^(sin u) / u, which is not f'(u). This means the problem statement as written leads to an inconsistency or a much more complex problem than usually intended for these exams.

Comparing with the "similar question" provided, which has d/dx F(x) = e^(sin x)/x, it is highly probable that the given question has a typo and was intended to be d/dx(f(x)) = e^(sin x) / x. Let's proceed with this assumption, as it makes the problem solvable and consistent.

Assumption: Let d/dx(f(x)) = e^(sin x) / x. This means f'(x) = e^(sin x) / x.

Under this assumption, the integrand (e^(sin u) / u) is exactly f'(u). By the Fundamental Theorem of Calculus, ∫_{1}^{16} f'(u) du = f(16) - f(1).

The problem states that ∫_{1}^{4} 2 \frac{e^{sinx^2}}{x}dx = F(K)-f(1). Assuming F(K) refers to f(K) (as f(1) is used), we have: f(16) - f(1) = f(K) - f(1)

Comparing both sides, we can conclude that f(16) = f(K). Since f'(x) = e^(sin x) / x and for x > 0, e^(sin x) is always positive and x is positive, f'(x) is always positive. This means f(x) is a strictly increasing function. Therefore, if f(16) = f(K), then K must be 16.

The final answer is $\boxed{\text{16}}$ .

Explanation of the solution:

Identify the derivative: Assume the intended derivative is f'(x) = e^(sin x) / x based on common problem patterns and the provided similar question.
Transform the integral: Use the substitution u = x^2. This implies du = 2x dx.
Adjust integrand and limits: Rewrite the integral ∫_{1}^{4} 2 \frac{e^{sinx^2}}{x}dx as ∫_{1}^{4} \frac{e^{sinx^2}}{x^2} (2x dx).
Substitute: Replace x^2 with u and 2x dx with du. Change limits from x=1 to u=1, and x=4 to u=16. The integral becomes ∫_{1}^{16} \frac{e^{sin u}}{u} du.
Apply Fundamental Theorem of Calculus: Since f'(u) = e^(sin u) / u, the integral evaluates to [f(u)]_{1}^{16} = f(16) - f(1).
Compare and solve for K: Equate this result with the given expression f(K) - f(1). This yields f(16) - f(1) = f(K) - f(1), which simplifies to f(16) = f(K). Since f'(x) > 0 for x > 0, f(x) is strictly increasing, implying K = 16.